A Rearranging a Sequence
#include<bits/stdc++.h>
using namespace std;
const double eps=1e-10;
const double pi=3.1415926535897932384626433832795;
const double eln=2.718281828459045235360287471352;
#define LL long long
#define IN freopen("in.txt", "r", stdin)
#define OUT freopen("out.txt", "w", stdout)
#define scan(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define sqr(x) (x) * (x)
#define pr(x) printf("Case %d: ",x)
#define prn(x) printf("Case %d:\n",x)
#define prr(x) printf("Case #%d: ",x)
#define prrn(x) printf("Case #%d:\n",x)
#define lowbit(x) (x&(-x))
#define fi first
#define se second
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;
const int maxn=200005;
int n,m;
int a[maxn];
bool vis[maxn];
int main()
{
scanf("%d%d",&n,&m);
for(int i=1;i<=m;i++)scanf("%d",&a[i]);
int cnt=0;
for(int i=m;i>=1;i--)
{
if(vis[a[i]])continue;
printf("%d\n",a[i]);
vis[a[i]]=true;
}
for(int i=1;i<=n;i++)
{
if(vis[i])continue;
cnt++;
printf("%d\n",i);
}
return 0;
}
B Quality of Check Digits
#include<bits/stdc++.h>
using namespace std;
const double eps=1e-10;
const double pi=3.1415926535897932384626433832795;
const double eln=2.718281828459045235360287471352;
#define LL long long
#define IN freopen("in.txt", "r", stdin)
#define OUT freopen("out.txt", "w", stdout)
#define scan(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define sqr(x) (x) * (x)
#define pr(x) printf("Case %d: ",x)
#define prn(x) printf("Case %d:\n",x)
#define prr(x) printf("Case #%d: ",x)
#define prrn(x) printf("Case #%d:\n",x)
#define lowbit(x) (x&(-x))
#define fi first
#define se second
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;
int a[10][10];
bool vis[10];
int x[4],v[4],u[4],cnt=0;
int check(int q,int w,int e,int r)
{
u[0]=q;u[1]=w;u[2]=e;u[3]=r;
int ret=0;
for(int i=0;i<4;i++)
ret=a[ret][u[i]];
return ret;
}
int main()
{
for(int i=0;i<=9;i++)
for(int j=0;j<=9;j++)
scanf("%d",&a[i][j]);
for(int i=0;i<=9;i++)
for(int j=0;j<=9;j++)
if(i!=j && a[i][j]==0)vis[i]=true;
for(x[0]=0;x[0]<=9;x[0]++)
for(x[1]=0;x[1]<=9;x[1]++)
for(x[2]=0;x[2]<=9;x[2]++)
for(x[3]=0;x[3]<=9;x[3]++)
{
int now=check(x[0],x[1],x[2],x[3]);
if(vis[now]){cnt++;continue;}
bool f=true;
for(int p=0;p<4 && f;p++)
{
for(int t=0;t<4;t++)v[t]=x[t];
for(int s=0;s<=9;s++)
if(s!=x[p])
{
v[p]=s;
if(a[check(v[0],v[1],v[2],v[3])][now]==0)
{
f=false;break;
}
}
}
if(!f){cnt++;continue;}
if(x[0]!=x[1] && a[check(x[1],x[0],x[2],x[3])][now]==0){cnt++;continue;}
if(x[1]!=x[2] && a[check(x[0],x[2],x[1],x[3])][now]==0){cnt++;continue;}
if(x[2]!=x[3] && a[check(x[0],x[1],x[3],x[2])][now]==0){cnt++;continue;}
if(x[3]!=now && a[check(x[0],x[1],x[2],now)][x[3]]==0){cnt++;continue;}
}
printf("%d\n",cnt);
return 0;
}
C Distribution Center
可以发现到达某个点的点一定是一段区间，只要求上下界。
那么可以考虑相邻2 Lanes的转移，每个Lane存它能到达的上、下界。
#include<bits/stdc++.h>
using namespace std;
const double eps=1e-10;
const double pi=3.1415926535897932384626433832795;
const double eln=2.718281828459045235360287471352;
#define LL long long
#define IN freopen("in.txt", "r", stdin)
#define OUT freopen("out.txt", "w", stdout)
#define scan(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define sqr(x) (x) * (x)
#define pr(x) printf("Case %d: ",x)
#define prn(x) printf("Case %d:\n",x)
#define prr(x) printf("Case #%d: ",x)
#define prrn(x) printf("Case #%d:\n",x)
#define lowbit(x) (x&(-x))
#define fi first
#define se second
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;
int n,m;
vector<int> g[200005];
int a[2][100005];
int up[200005],dw[200005];
int main()
{
//IN;
scanf("%d%d",&n,&m);
for(int i=0;i<=n;i++)g[i].pb(0),g[i].pb(100005);
for(int i=1;i<=m;i++)
{
int x,y;
scanf("%d%d",&x,&y);
g[y].pb(x);
}
for(int i=1;i<=n;i++)up[i]=dw[i]=i;
for(int i=1;i<n;i++)sort(g[i].begin(),g[i].end());
up[1]=1;
int now=0;
for(int i=1;i<n;i++)
{
now^=1;
a[now^1][0]=i;
for(int t : g[i])
{
if(t<=0 || t>=100000)continue;
int x=lower_bound(g[i-1].begin(),g[i-1].end(),t)-g[i-1].begin();
x--;
a[now][t]=a[now^1][g[i-1][x]];
up[i+1]=min(up[i+1],a[now][t]);
}
}
now=0;dw[n]=n;a[0][0]=n;
for(int i=n-1;i>=1;i--)
{
now^=1;
a[now][0]=i;
for(int t : g[i])
{
if(t<=0 || t>=100000)continue;
int x=lower_bound(g[i+1].begin(),g[i+1].end(),t)-g[i+1].begin();
x--;
a[now][t]=a[now^1][g[i+1][x]];
dw[i]=max(dw[i],a[now][t]);
}
}
for(int i=1;i<=n;i++)printf("%d%c",dw[i]-up[i]+1,i==n?'\n':' ');
return 0;
}
D Hidden Anagrams
双hash
#include<bits/stdc++.h>
using namespace std;
const double eps=1e-10;
const double pi=3.1415926535897932384626433832795;
const double eln=2.718281828459045235360287471352;
#define LL long long
#define IN freopen("in.txt", "r", stdin)
#define OUT freopen("out.txt", "w", stdout)
#define scan(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define sqr(x) (x) * (x)
#define pr(x) printf("Case %d: ",x)
#define prn(x) printf("Case %d:\n",x)
#define prr(x) printf("Case #%d: ",x)
#define prrn(x) printf("Case #%d:\n",x)
#define lowbit(x) (x&(-x))
#define fi first
#define se second
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;
const int mod1=1e9+7;
const int seed1=5003;
const int mod2=1e9+9;
const int seed2=5009;
char s1[4005],s2[4005];
int a1[4005][26],a2[4005][26];
int l1,l2;
pii h1[4005],h2[4005];
int c1,c2;
int main()
{
scanf("%s",s1+1);l1=strlen(s1+1);
scanf("%s",s2+1);l2=strlen(s2+1);
for(int i=1;i<=l1;i++)
{
for(int j=0;j<26;j++)
a1[i][j]=a1[i-1][j];
a1[i][s1[i]-97]++;
}
for(int i=1;i<=l2;i++)
{
for(int j=0;j<26;j++)
a2[i][j]=a2[i-1][j];
a2[i][s2[i]-97]++;
}
for(int ans=min(l1,l2);ans>=1;ans--)
{
c1=c2=0;
for(int i=1;i+ans-1<=l1;i++)
{
int now1=0,now2=0;
for(int j=0;j<26;j++)
{
int p=a1[i+ans-1][j]-a1[i-1][j];
now1=1ll*now1*seed1%mod1;
now1+=p;
now1%=mod1;
now2+=p;now2%=mod2;
now2=1ll*now2*seed2%mod2;
}
h1[++c1]=mp(now1,now2);
}
for(int i=1;i+ans-1<=l2;i++)
{
int now1=0,now2=0;
for(int j=0;j<26;j++)
{
int p=a2[i+ans-1][j]-a2[i-1][j];
now1=1ll*now1*seed1%mod1;
now1+=p;
now1%=mod1;
now2+=p;now2%=mod2;
now2=1ll*now2*seed2%mod2;
}
h2[++c2]=mp(now1,now2);
}
sort(h1+1,h1+c1+1);
sort(h2+1,h2+c2+1);
int l=1;
for(int i=1;i<=c1;i++)
{
while(l<=c2 && h2[l]<h1[i])l++;
if(h1[i]==h2[l])
{
printf("%d\n",ans);
return 0;
}
}
}
printf("0\n");
return 0;
}
E Infallibly Crack Perplexing Cryptarithm
逆波兰表达式，递归下降
F Three Kingdoms of Bourdelot
假设条目可以全是negative，任意两人没有血缘关系。
维护一个队列，表示一定成立的条目。先将第一条目入队，之后找出可以确定的条目，继续入队……反复执行此操作，如果存在传递闭包则无解。
G Placing Medals on a Binary Tree
我们考虑对第i层的一个节点占用了(1/2)i的总节点。
于是我们实际只要判断之前用的节点+当前节点占总节点的比值是否>1。
注意这题精度要求较高，我们可以用线段树，树状数组维护大量(1/2)k相加，相当于手动模拟高精加运算。注意每次如果不进位则单点+1，进位则若干为变为0，单点+1，容易发现变为0这个操作最多发生O(n)次，所以可以手动模拟。
考虑最劣情况1/2,1/4,1/8,..., \frac 1 {2^{n-1}} , \frac 1 {2^{n-1}
或者1/2k,1/2k,...,1/2k
发现最多会影响的位置是第n+log(n)层
所以超过这层的放在这层就行
#include<bits/stdc++.h>
using namespace std;
const double eps=1e-10;
const double pi=3.1415926535897932384626433832795;
const double eln=2.718281828459045235360287471352;
#define LL long long
#define IN freopen("in.txt", "r", stdin)
#define OUT freopen("out.txt", "w", stdout)
#define scan(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define sqr(x) (x) * (x)
#define pr(x) printf("Case %d: ",x)
#define prn(x) printf("Case %d:\n",x)
#define prr(x) printf("Case #%d: ",x)
#define prrn(x) printf("Case #%d:\n",x)
#define lowbit(x) (x&(-x))
#define fi first
#define se second
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;
int a[600005],s[600005];
int n,x,cs=555555;
void add(int p,int nu)
{
while(p<=cs)
{
s[p]+=nu;
p+=lowbit(p);
}
}
int cal(int p)
{
int ret=0;
while(p>0)
{
ret+=s[p];
p-=lowbit(p);
}
return ret;
}
int main()
{
scanf("%d",&n);
while(n--)
{
scanf("%d",&x);
x=min(x,cs);
if(a[0]>0 || (cal(x)==x && cal(cs)>cal(x)))
{
printf("No\n");
continue;
}
printf("Yes\n");
while(a[x]==1)
{
a[x]=0;
add(x,-1);
x--;
}
a[x]=1;
if(x>0)add(x,1);
}
return 0;
}
H Animal Companion in Maze
把问题转化成给一颗带点权的树(2 ≤ n ≤ 10000），对每个点A求maxB(dis(A,B)+valueB)
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
#define MAXN 1000001
int dfn[MAXN], low[MAXN], stack[MAXN];
int res[MAXN];
bool flag[MAXN];
vector<int> e[MAXN];
int cur, top, cnt;//cur为访问标号，top为栈顶，cnt为强连通分量个数
//使用前清空dfn,flag数组，cur，cnt置0；返回res存储每个点所在强连通分量集合
void tarjan(int i)
{
low[i] = dfn[i] = ++cur;
stack[++top] = i; flag[i] = true;
for (unsigned j = 0; j < e[i].size(); j++){
int id = e[i][j];
if (!dfn[id]){
tarjan(id);
low[i] = min(low[i], low[id]);
}
else if (flag[id])
low[i] = min(low[i], dfn[id]);//此处可写low
}
if (low[i] == dfn[i]){
cnt++;
int t;
do{
t = stack[top--];
res[t] = cnt;
flag[t] = false;
} while (t != i);
}
}
#include<cstdio>
#include<vector>
#include<cstring>
using namespace std;
vector<int> v[1000001];
int degree[1000001];
int ans[1000001], layer[1000001];//ans为拓扑序列，layer为每个结点层数
//返回值为true表示成功，false表示有环
bool toposort(int n)
{
int k = 0;
memset(degree + 1, 0, sizeof(int)*n);
memset(layer + 1, 0, sizeof(int)*n);
for (int i = 1; i <= n; i++){
for (unsigned int t = 0; t < v[i].size(); t++)
degree[v[i][t]]++;
}
for (int i = 1; i <= n; i++){
if (!degree[i])ans[k++] = i;
}
for (int j = 0; j < k; j++){
for (unsigned int t = 0; t < v[ans[j]].size(); t++){
int i = v[ans[j]][t];
if (!--degree[i]){
ans[k++] = i;
layer[i] = layer[ans[j]] + 1;
}
}
}
return k == n;
}
vector<int> vv[200001];
vector<int> s[200001];
int f[200001];
bool vis[200001];
int getFather(int i){
if(!f[i])return i;
return f[i]=getFather(f[i]);
}
struct Edge{
int x,y,w;
}ee[200001];
int out[200001];
int dp[200001],dp2[200001],dp3[200001],son[200001];
void dfs1(int i,int fa)
{
dp2[i]=dp[i];
for(int j:vv[i]){
if(j!=fa){
dfs1(j,i);
int t=dp[j]+1;
if(t>dp[i]){dp2[i]=dp[i];dp[i]=t;son[i]=j;}
else if(t>dp2[i]){dp2[i]=t;}
}
}
}
void dfs2(int i,int fa)
{
for(int j:vv[i]){
if(j!=fa){
if(j==son[i])dp3[j]=dp2[i]+1;
else dp3[j]=dp[i]+1;
dp3[j]=max(dp3[j],dp3[i]+1);
dfs2(j,i);
}
}
}
int main()
{
int n,m;
scanf("%d%d",&n,&m);
bool flag=false;
for(int i=0;i<m;i++){
scanf("%d%d%d",&ee[i].x,&ee[i].y,&ee[i].w);
if(ee[i].w==2){
int x=getFather(ee[i].x);
int y=getFather(ee[i].y);
if(x==y){flag=true;break;}
f[x]=y;
e[ee[i].x].push_back(ee[i].y);
e[ee[i].y].push_back(ee[i].x);
}
else e[ee[i].x].push_back(ee[i].y);
}
if(!flag){
for(int i=1;i<=n;i++){
if(!dfn[i])tarjan(i);
}
for(int i=0;i<m;i++){
if(res[ee[i].x]==res[ee[i].y]&&ee[i].w==1){
flag=true;break;
}
}
}
if(flag)printf("Infinite");
else{
int finally=0;
for(int i=1;i<=n;i++){
for(int j:e[i]){
if(res[i]!=res[j])
v[res[i]].push_back(res[j]);
else
vv[i].push_back(j);
}
}
toposort(cnt);
for(int i=1;i<=n;i++)
s[res[i]].push_back(i);
for(int i=cnt-1;i>=0;i--){
int ii=ans[i];
for(int j:s[ii]){
for(int k:e[j]){
if(res[k]!=ii)
dp[j]=max(dp[j],dp[k]+1);
}
}
dfs1(s[ii][0],0);
dfs2(s[ii][0],0);
for(int j:s[ii]){
dp[j]=max(dp[j],dp3[j]);
finally=max(finally,max(dp[j],dp3[j]));
}
}
printf("%d",finally);
}
}
I Skinny Polygon
数论
J Cover the Polygon with Your Di
求一个多边形和圆相交的最大值
爬山
#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b)
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
ll sqr(ll a){return a*a;}
ld sqr(ld a){return a*a;}
double sqr(double a){return a*a;}
const double eps=1e-10;
int dcmp(double x) {
if (fabs(x)<eps) return 0; else return x<0 ? -1 : 1;
}
ld PI = 3.141592653589793238462643383;
class P{
public:
double x,y;
P(double x=0,double y=0):x(x),y(y){}
friend ld dis2(P A,P B){return sqr(A.x-B.x)+sqr(A.y-B.y); }
friend ld Dot(P A,P B) {return A.x*B.x+A.y*B.y; }
friend ld Length(P A) {return sqrt(Dot(A,A)); }
friend ld Angle(P A,P B) {
if (dcmp(Dot(A,A))==0||dcmp(Dot(B,B))==0||dcmp(Dot(A-B,A-B))==0) return 0;
return acos(max((ld)-1.0, min((ld)1.0, Dot(A,B) / Length(A) / Length(B) )) );
}
friend P operator- (P A,P B) { return P(A.x-B.x,A.y-B.y); }
friend P operator+ (P A,P B) { return P(A.x+B.x,A.y+B.y); }
friend P operator* (P A,double p) { return P(A.x*p,A.y*p); }
friend P operator/ (P A,double p) { return P(A.x/p,A.y/p); }
friend bool operator< (const P& a,const P& b) {return dcmp(a.x-b.x)<0 ||(dcmp(a.x-b.x)==0&& dcmp(a.y-b.y)<0 );}
};
P read_point() {
P a;
scanf("%lf%lf",&a.x,&a.y);
return a;
}
bool operator==(const P& a,const P& b) {
return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y) == 0;
}
typedef P V;
double Cross(V A,V B) {return A.x*B.y - A.y*B.x;}
double Area2(P A,P B,P C) {return Cross(B-A,C-A);}
V Rotate(V A,double rad) {
return V(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));
}
// A 不是 0向量
V Normal(V A) {
double L = Length(A);
return V(-A.y/L , A.x/L);
}
namespace complex_G{
typedef complex<double> Point;
//real(p):实部 imag(p):虚部 conj(p):共轭
typedef Point Vector;
double Dot(Vector A,Vector B) {return real(conj(A)*B); }
double Cross(Vector A,Vector B) {return imag(conj(A)*B); }
Vector Rotate(Vector A,double rad) {return A*exp(Point(0,rad)); }
}
//Cross(v,w)==0(平行)时,不能调这个函数
P GetLineIntersection(P p,V v,P Q,V w){
V u = p-Q;
double t = Cross(w,u)/Cross(v,w);
return p+v*t;
}
P GetLineIntersectionB(P p,V v,P Q,V w){
return GetLineIntersection(p,v-p,Q,w-Q);
}
double DistanceToLine(P p,P A,P B) {
V v1 = B-A, v2 = p-A;
return fabs(Cross(v1,v2))/Length(v1);
}
double DistanceToSegment(P p,P A,P B) {
if (A==B) return Length(p-A);
V v1 = B-A, v2 = p-A, v3 = p - B;
if (dcmp(Dot(v1,v2))<0) return Length(v2);
else if (dcmp(Dot(v1,v3))>0 ) return Length(v3);
else return fabs(Cross(v1,v2) ) / Length(v1);
}
P GetLineProjection(P p,P A,P B) {
V v=B-A;
return A+v*(Dot(v,p-A)/Dot(v,v));
}
//规范相交-线段相交且交点不在端点
bool SegmentProperIntersection(P a1,P a2,P b1,P b2) {
double c1 = Cross(a2-a1,b1-a1) , c2 = Cross(a2-a1,b2-a1),
c3 = Cross(b2-b1,a1-b1) , c4 = Cross(b2-b1,a2-b1);
return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0;
}
//点在线段上（不包含端点）
bool OnSegment(P p,P a1,P a2) {
return dcmp(Cross(a1-p,a2-p)) == 0 && dcmp(Dot(a1-p,a2-p))<0;
}
double PolygonArea(P *p,int n) {
double area=0;
For(i,n-2) area+=Cross(p[i]-p[0],p[i+1]-p[0]);
return area/2;
}
/*欧拉公式： V+F-E=2
V-点数 F面数 E边数 */
struct C{
P c;
double r,x,y;
C(P c,double r):c(c),r(r),x(c.x),y(c.y){}
P point(double a) {
return P(c.x+cos(a)*r,c.y+sin(a)*r);
}
};
struct Line{
P p;
V v;
double ang;
Line(){}
Line(P p,V v):p(p),v(v) {ang=atan2(v.y,v.x); }
bool operator<(const Line & L) const {
return ang<L.ang;
}
P point(double a) {
return p+v*a;
}
};
int getLineCircleIntersection(Line L,C cir,double &t1,double &t2,vector<P> & sol) {
if (dcmp(DistanceToLine(cir.c,L.p,L.p+L.v)-cir.r)==0) {
P A=GetLineProjection(cir.c,L.p,L.p+L.v);
sol.pb(A);
t1 = (A-L.p).x / L.v.x;
return 1;
}
double a = L.v.x, b = L.p.x - cir.c.x, c = L.v.y, d= L.p.y - cir.c.y;
double e = a*a+c*c, f = 2*(a*b + c*d), g = b*b+d*d-cir.r*cir.r;
double delta = f*f - 4*e*g;
if (dcmp(delta)<0) return 0;
else if (dcmp(delta)==0) {
t1 = t2 = -f / (2*e); sol.pb(L.point(t1));
return 1;
}
t1 = (-f - sqrt(delta)) / (2*e); sol.pb(L.point(t1));
t2 = (-f + sqrt(delta)) / (2*e); sol.pb(L.point(t2));
return 2;
}
double angle(V v) {return atan2(v.y,v.x);}
int getCircleCircleIntersection(C C1,C C2,vector<P>& sol) {
double d = Length(C1.c-C2.c);
if (dcmp(d)==0) {
if (dcmp(C1.r - C2.r)==0) return -1; //2圆重合
return 0;
}
if (dcmp(C1.r+C2.r-d)<0) return 0;
if (dcmp(fabs(C1.r-C2.r)-d)>0) return 0;
double a = angle(C2.c-C1.c);
double da = acos((C1.r*C1.r+d*d - C2.r*C2.r)/ (2*C1.r*d));
P p1 = C1.point(a-da), p2 = C1.point(a+da);
sol.pb(p1);
if (p1==p2) return 1;
sol.pb(p2);
return 2;
}
// Tangents-切线
int getTangents(P p,C c,V* v) {
V u= c.c-p;
double dist = Length(u);
if (dist<c.r) return 0;
else if (dcmp(dist-c.r)==0) {
v[0]=Rotate(u,PI/2);
return 1;
} else {
double ang = asin(c.r / dist);
v[0]=Rotate(u,-ang);
v[1]=Rotate(u,ang);
return 2;
}
}
int getTangentsPoint(P p,C c,P *point) {
V u= c.c-p;
double dist = Length(u);
if (dist<c.r) return 0;
else if (dcmp(dist-c.r)==0) {
point[0]=p;
return 1;
} else {
V v[2];
double ang = asin(c.r / dist);
v[0]=Rotate(u,-ang);point[0]=GetLineProjection(c.c,p,p+v[0]);
v[1]=Rotate(u,ang); point[1]=GetLineProjection(c.c,p,p+v[1]);
return 2;
}
}
//这个函数假设整数坐标和整数半径
//double时要把int改成double
int getTangents(C A,C B,P* a,P* b) {
int cnt=0;
if (A.r<B.r) {swap(A,B),swap(a,b);}
int d2 = (A.c.x-B.c.x)*(A.c.x-B.c.x) + (A.c.y-B.c.y)*(A.c.y-B.c.y);
int rdiff = A.r-B.r;
int rsum = A.r+B.r;
if (d2<rdiff*rdiff) return 0;
double base = atan2(B.y-A.y,B.x-A.x);
if (d2==0 && A.r == B.r) return -1;
if (d2 == rdiff*rdiff) {
a[cnt] = A.point(base); b[cnt] = B.point(base); ++cnt;
return 1;
}
double ang = acos((A.r-B.r)/sqrt(d2));
a[cnt] = A.point(base+ang); b[cnt] = B.point(base+ang); ++cnt;
a[cnt] = A.point(base-ang); b[cnt] = B.point(base-ang); ++cnt;
if (d2==rsum*rsum) {
a[cnt] = A.point(base); b[cnt] = B.point(PI+base); ++cnt;
}
else if (d2>rsum*rsum) {
double ang = acos((A.r+B.r)/sqrt(d2));
a[cnt] = A.point(base+ang); b[cnt] = B.point(PI+base+ang); ++cnt;
a[cnt] = A.point(base-ang); b[cnt] = B.point(PI+base-ang); ++cnt;
}
return cnt;
}
//Circumscribed-外接
C CircumscribedCircle(P p1,P p2,P p3) {
double Bx = p2.x-p1.x, By= p2.y-p1.y;
double Cx = p3.x-p1.x, Cy= p3.y-p1.y;
double D = 2*(Bx*Cy-By*Cx);
double cx = (Cy*(Bx*Bx+By*By)-By*(Cx*Cx+Cy*Cy))/D + p1.x;
double cy = (Bx*(Cx*Cx+Cy*Cy)-Cx*(Bx*Bx+By*By))/D + p1.y;
P p =P(cx,cy);
return C(p,Length(p1-p));
}
//Inscribed-内接
C InscribedCircle(P p1,P p2,P p3) {
double a = Length(p2-p3);
double b = Length(p3-p1);
double c = Length(p1-p2);
P p = (p1*a+p2*b+p3*c)/(a+b+c);
return C(p,DistanceToLine(p,p1,p2));
}
double torad(double deg) {
return deg/180*acos(-1);
}
//把 角度+-pi 转换到 [0,pi) 上
double radToPositive(double rad) {
if (dcmp(rad)<0) rad=ceil(-rad/PI)*PI+rad;
if (dcmp(rad-PI)>=0) rad-=floor(rad/PI)*PI;
return rad;
}
double todeg(double rad) {
return rad*180/acos(-1);
}
//(R,lat,lng)->(x,y,z)
void get_coord(double R,double lat,double lng,double &x,double &y,double &z) {
lat=torad(lat);
lng=torad(lng);
x=R*cos(lat)*cos(lng);
y=R*cos(lat)*sin(lng);
z=R*sin(lat);
}
void print(double a) {
printf("%.6lf",a);
}
void print(P p) {
printf("(%.6lf,%.6lf)",p.x,p.y);
}
template<class T>
void print(vector<T> v) {
sort(v.begin(),v.end());
putchar('[');
int n=v.size();
Rep(i,n) {
print(v[i]);
if (i<n-1) putchar(',');
}
puts("]");
}
// 把直线沿v平移
// Translation-平移
Line LineTranslation(Line l,V v) {
l.p=l.p+v;
return l;
}
void CircleThroughAPointAndTangentToALineWithRadius(P p,Line l,double r,vector<P>& sol) {
V e=Normal(l.v);
Line l1=LineTranslation(l,e*r),l2=LineTranslation(l,e*(-r));
double t1,t2;
getLineCircleIntersection(l1,C(p,r),t1,t2,sol);
getLineCircleIntersection(l2,C(p,r),t1,t2,sol);
}
void CircleTangentToTwoLinesWithRadius(Line l1,Line l2,double r,vector<P>& sol) {
V e1=Normal(l1.v),e2=Normal(l2.v);
Line L1[2]={LineTranslation(l1,e1*r),LineTranslation(l1,e1*(-r))},
L2[2]={LineTranslation(l2,e2*r),LineTranslation(l2,e2*(-r))};
Rep(i,2) Rep(j,2) sol.pb(GetLineIntersection(L1[i].p,L1[i].v,L2[j].p,L2[j].v));
}
void CircleTangentToTwoDisjointCirclesWithRadius(C c1,C c2,double r,vector<P>& sol) {
c1.r+=r; c2.r+=r;
getCircleCircleIntersection(c1,c2,sol);
}
//确定4个点能否组成凸多边形，并按顺序（不一定是逆时针）返回
bool ConvexPolygon(P &A,P &B,P &C,P &D) {
if (SegmentProperIntersection(A,C,B,D)) return 1;
swap(B,C);
if (SegmentProperIntersection(A,C,B,D)) return 1;
swap(D,C);
if (SegmentProperIntersection(A,C,B,D)) return 1;
return 0;
}
bool IsParallel(P A,P B,P C,P D) {
return dcmp(Cross(B-A,D-C))==0;
}
bool IsPerpendicular(V A,V B) {
return dcmp(Dot(A,B))==0;
}
//先调用ConvexPolygon 求凸包并确认是否是四边形
// Trapezium-梯形 Rhombus-菱形
bool IsTrapezium(P A,P B,P C,P D){
return IsParallel(A,B,C,D)^IsParallel(B,C,A,D);
}
bool IsParallelogram(P A,P B,P C,P D) {
return IsParallel(A,B,C,D)&&IsParallel(B,C,A,D);
}
bool IsRhombus(P A,P B,P C,P D) {
return IsParallelogram(A,B,C,D)&&dcmp(Length(B-A)-Length(C-B))==0;
}
bool IsRectangle(P A,P B,P C,P D) {
return IsParallelogram(A,B,C,D)&&IsPerpendicular(B-A,D-A);
}
bool IsSquare(P A,P B,P C,P D) {
return IsParallelogram(A,B,C,D)&&IsPerpendicular(B-A,D-A)&&dcmp(Length(B-A)-Length(C-B))==0;
}
bool IsCollinear(P A,P B,P C,P D) { //共线
return !dcmp(Cross(A-B,C-D)) && !dcmp(Cross(A-B,C-B)) && !dcmp(DistanceToLine(A,C,D));
}
//chord-弦 arc-弧
double ArcDis(double chord,double r) {
return 2*asin(chord/2/r)*r;
}
typedef vector<P> Polygon ;
int isPointInPolygon(P p,Polygon poly) {
int wn=0;
int n=poly.size();
Rep(i,n) {
if (OnSegment(p,poly[i],poly[(i+1)%n])) return -1; //edge
int k=dcmp(Cross(poly[(i+1)%n]-poly[i],p-poly[i]));
int d1 = dcmp(poly[i].y-p.y);
int d2 = dcmp(poly[(i+1)%n].y-p.y);
if ( k > 0 && d1 <= 0 && d2 > 0 ) wn++;
if ( k < 0 && d2 <= 0 && d1 > 0 ) wn--;
}
if (wn!=0) return 1; //inside
return 0; //outside
}
int ConvexHull(P *p,int n,P *ch) {
sort(p,p+n);
int m=0;
Rep(i,n) {
while(m>1 && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
ch[m++]=p[i];
}
int k=m;
RepD(i,n-2) {
while(m>k && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
ch[m++]=p[i];
}
if ( n > 1 ) m--;
return m;
}
//把两点式转为一般式 ax+by+c=0
void Two_pointFormToGeneralForm(P A,P B,double &a,double &b,double &c){
a = A.y - B.y;
b = B.x - A.x;
c = Cross(A,B);
}
//有向直线A->B 切割多边行poly 可能返回单点线段 O(n)
Polygon CutPolygon(Polygon poly,P A,P B){
Polygon newpoly;
int n=poly.size();
Rep(i,n) {
P C = poly[i];
P D = poly[(i+1)%n];
if (dcmp(Cross(B-A,C-A))>=0) newpoly.pb(C);
if (dcmp(Cross(B-A,C-D))) {
P ip = GetLineIntersection(A,B-A,C,D-C);
if (OnSegment(ip,C,D)) newpoly.pb(ip);
}
}
return newpoly;
}
double PolygonArea(Polygon &p) {
double area=0;
int n=p.size();
For(i,n-2) area+=Cross(p[i]-p[0],p[i+1]-p[0]);
return area/2;
}
//线上不算
bool OnLeft(Line L,P p) {
return Cross(L.v,p-L.p)>0;
}
P GetIntersection(Line a,Line b) {
V u=a.p-b.p;
double t = Cross(b.v,u) / Cross(a.v, b.v);
return a.p + a.v*t;
}
int HalfplaneIntersection(Line *L, int n, P* poly) {
sort(L,L+n);
int fi,la;
P *p = new P[n];
Line *q = new Line[n];
q[fi=la=0 ] = L[0];
For(i,n-1) {
while (fi < la && !OnLeft(L[i],p[la-1])) la--;
while (fi < la && !OnLeft(L[i],p[fi])) fi++;
q[++la] = L[i];
if (fabs(Cross(q[la].v, q[la-1].v))<eps) {
la--;
if (OnLeft(q[la],L[i].p)) q[la] = L[i];
}
if (fi<la) p[la-1] = GetIntersection(q[la-1],q[la]);
}
while(fi < la && !OnLeft(q[fi],p[la-1])) la--;
if (la-fi<=1) return 0;
p[la] = GetIntersection(q[la],q[fi]);
int m=0;
Fork(i,fi,la) poly[m++]=p[i];
return m;
}
// hypot(double x,double y); return sqrt(x*x+y*y)
double rotating_calipers(P *p,int n) {
int q=1;
double ans=0;
Rep(i,n) {
while(fabs(Area2(p[i],p[(i+1)%n],p[(q+1)%n]))> fabs(Area2(p[i],p[(i+1)%n],p[q] )) )
q=(q+1)%n;
ans = max( ans, (double)max(Length(p[i]- p[q]), Length(p[(i+1)%n] -p[q] ) ) );
}
return ans;
}
// rotating for rect
double rec_rotating_calipers(P *p,int n) {
int q=1;
double ans1=1e15,ans2=1e15;
int l=0,r=0;
Rep(i,n) {
while(dcmp(fabs(Area2(p[i],p[(i+1)%n],p[(q+1)%n])) - fabs(Area2(p[i],p[(i+1)%n],p[q] ))) >0 )
q=(q+1)%n;
while (dcmp(Dot(p[(i+1)%n]-p[i],p[(r+1)%n]-p[r]))>0 ) r=(r+1)%n;
if (!i) l=q;
while (dcmp(Dot(p[(i+1)%n]-p[i],p[(l+1)%n]-p[l]))<0 ) l=(l+1)%n;
double d = Length(p[(i+1)%n]-p[i]);
double h = fabs(Area2(p[i],p[(i+1)%n],p[q] ))/d;
double w=(Dot(p[(i+1)%n]-p[i],p[r]-p[i]) - Dot(p[(i+1)%n]-p[i],p[l]-p[i]) ) /d;
ans1=min(ans1,2*(h+w)),ans2=min(ans2,h*w);
}
printf("%.2lf %.2lf\n",ans2,ans1);
}
//去共线化
void simplify(Polygon& poly) {
Polygon ans;
int n=SI(poly);
Rep(i,n) {
if (dcmp(Cross(poly[i]-poly[(i+1)%n],poly[(i+1)%n]-poly[(i+2)%n]))!=0)
ans.pb(poly[(i+1)%n]);
}
n=SI(ans);
cout<<n<<endl;
Rep(i,n) printf("%.4lf %.4lf\n",ans[i].x,ans[i].y);
}
int getSegCircleIntersection(Line L,C cir,vector<P> & sol) {
if (dcmp(DistanceToLine(cir.c,L.p,L.p+L.v)-cir.r)==0) {
P A= GetLineProjection(cir.c,L.p,L.p+L.v);
if (OnSegment(A,L.p,L.p+L.v) || L.p==A || L.p+L.v==A )
sol.pb(A);
return sol.size();
}
double t1,t2;
double a = L.v.x, b = L.p.x - cir.c.x, c = L.v.y, d= L.p.y - cir.c.y;
double e = a*a+c*c, f = 2*(a*b + c*d), g = b*b+d*d-cir.r*cir.r;
double delta = f*f - 4*e*g;
if (dcmp(delta)<0) return 0;
else if (dcmp(delta)==0) {
t1 = -f / (2*e);
if (dcmp(t1)>=0&&dcmp(t1-1)<=0) {
sol.pb(L.point(t1));
}
return sol.size();
}
t1 = (-f - sqrt(delta)) / (2*e); if (dcmp(t1)>=0&&dcmp(t1-1)<=0) sol.pb(L.point(t1));
t2 = (-f + sqrt(delta)) / (2*e); if (dcmp(t2)>=0&&dcmp(t2-1)<=0) sol.pb(L.point(t2));
if(SI(sol)==2 && t1>t2) swap(sol[1],sol[0]);
return sol.size();
}
int isPointInOrOnCircle(P p,C c) {
return dcmp(Length(p-c.c)-c.r)<=0;
}
// Triangle(O,A,B) and Circle(O,m)
double CircleTriangleArea(P A,P B,double m) {
double ans=0;
C c=C(P(),m); P O=c.c;
if (A==O||B==O) return 0;
bool b = isPointInOrOnCircle(A,c);
bool b2 = isPointInOrOnCircle(B,c);
double opr;
if (dcmp(Area2(O,A,B))>=0) opr=1; else opr=-1;
if (b&&b2) {
ans+=opr*fabs(Area2(A,B,O))/2;
} else if (!b&&!b2){
Line l=Line(A,B-A);
vector<P> sol;
getSegCircleIntersection(l,c,sol);
if (SI(sol)==2) {
ans+=opr*fabs(Area2(sol[0],sol[1],O))/2;
ans+=opr*m*m/2*(Angle(A,sol[0])+Angle(sol[1],B));
} else {
ans+=opr*m*m/2*(Angle(A,B));
}
} else {
Line l=Line(A,B-A);
vector<P> sol;
getSegCircleIntersection(l,c,sol);
if (SI(sol)==2) {
ans+=opr*fabs(Area2(sol[0],sol[1],O))/2;
ans+=opr*m*m/2*(Angle(A,sol[0])+Angle(sol[1],B));
} else if(b) {
ans+=opr*fabs(Area2(sol[0],A,O))/2;
ans+=opr*m*m/2*(Angle(sol[0],B));
} else {
ans+=opr*fabs(Area2(sol[0],B,O))/2;
ans+=opr*m*m/2*(Angle(sol[0],A));
}
}
return ans;
}
double max_polygon_inside_segment(Polygon p) {
int n=SI(p);
double ans=0;
Rep(i,n) Fork(j,i+1,n-1) {
P A=p[i],B=p[j];
if (A==B) continue;
Polygon poly;
poly.pb(A); poly.pb(B);
Rep(k,n) {
P C=p[k],D=p[(k+1)%n];
if (IsParallel(A,B,C,D)) {
if (IsCollinear(A,B,C,D)) {
poly.pb(C);
poly.pb(D);
}
} else {
P t=GetLineIntersectionB(A,B,C,D);
poly.pb(t);
}
}
sort(ALL(poly));
poly.erase(unique(ALL(poly)),poly.end());
int sz=SI(poly);
double l=0; bool fl=0;
For(i,sz-1) {
P m=(poly[i]+poly[i-1])/2;
if (isPointInPolygon(m,p)) {
if(!fl) l=Length(poly[i]-poly[i-1]);
else l+=Length(poly[i]-poly[i-1]);
ans=max(ans,l);
fl=1;
}else fl=0;
}
}
return ans;
}
double Jinggai_problems(P *p,int n) {
int q=1;
double ans=INF;
Rep(i,n) {
while(fabs(Area2(p[i],p[(i+1)%n],p[(q+1)%n]))> fabs(Area2(p[i],p[(i+1)%n],p[q] )) )
q=(q+1)%n;
ans = min( ans, (double)DistanceToSegment(p[q],p[i],p[(i+1)%n]) );
}
return ans;
}
int n;
#define MAXN 912345
P a[MAXN];
double r;
double calc(P t) {
double s=0;
Rep(i,n) {
P A=a[i],B=a[(i+1)%n];
s+=CircleTriangleArea(A-t,B-t,r);
}
return s;
}
P d[4]={P(0,1),P(1,0),P(0,-1),P(-1,0)};
double solve() {
double T=100;
P o=P();
Rep(i,n) o=o+a[i]; o=o/n;
double ans=calc(o);
double eps=0.000001;
T=0;
Rep(i,n) Rep(j,n) gmax(T,(double)Length(a[i]-a[j]) );
while(T>eps)
{
double pre=ans;
double f=calc(o);
double fx=calc(o+P(eps,0));
double fy=calc(o+P(0,eps));
double fx2=(fx-f) / eps;
double fy2=(fy-f) /eps;
P v(fx2,fy2);
v=v/Length(v)*T;
P o2=o+v;
if (calc(o2)>calc(o)) {
o=o2;
}else T*=0.9;
}
return calc(o);
}
int main()
{
freopen("J.in","r",stdin);
n=read();
r;cin>>r;
Rep(i,n) a[i]=read_point();
cout<<setiosflags(ios::fixed)<<setprecision(10);
double ans=calc(a[1]);
P o=P();
Rep(i,n) o=o+a[i]; o=o/n;
printf("%.10lf\n",solve());
return 0;
}
K Black and White Boxes
超实数博弈
