G
Problem G. Gridlandia
Input file: standard input
Output file: standard output
Time limit: 1 second
Memory limit: 256 megabytes
给一个的矩阵，每个格子可以选取上下左右一条边（可以不选），所有选取的边不能共点。现在要求构造一个方案，选尽量多的边。
The continent of Gridlandia is a squares of side length n, divided into n2 square contries of side length. Each country has the resources to choose at most one of its four sides and build a ‘wall’ covering that entire side. However, since all the countries are at war, no two countries are willing to have their walls touch, even at the wall’s endpoints.
Find the maximum number of walls that can be built in Gridlandia, and construct a configuration where the maximum number of walls are built.
Input
The only line of input contains a single integer n (1 ≤ n ≤ 103) – the number of squares on each side of the grid.
Output
The output should consist of n lines.
Each line should contain n characters, each being blank (represented by ‘.’) or one of ‘ULDR’.
A dot represents a country with no walls, whereas each of the characters ‘ULDR’ represents a country that builds a wall on its ‘Up’, ‘Down’, ‘Left’, and ‘Right’ sides respectively.
#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 ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;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,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#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 Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(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)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline 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;
}
#define MAXN (1010)
char f[MAXN][MAXN];
int n;
bool inside(int i) {
return 1<=i&&i<=n;
}
bool inside(int i,int j) {
return inside(i)&inside(j);
}
bool ck(int i,int j ,int p) {
if(!inside(i,j) || f[i][j]=='.') return 1;
if(p==0) {
return (f[i][j]=='D')||(f[i][j]=='R');
}
if(p==1) {
return (f[i][j]=='D')||(f[i][j]=='L');
}
if(p==2) {
return (f[i][j]=='U')||(f[i][j]=='R');
}
if(p==3) {
return (f[i][j]=='U')||(f[i][j]=='L');
}
}
bool ck(int i,int j,int p,int p2) {
return ck(i,j,p)&&ck(i,j,p2);
}
int ct=0;
void dfs(int i,int j,int s=0) {
int id=n*(i-1)+j;
if(n*n-id+s<ct) return;
if(i==n+1 && j==1) {
// if(s==9) {
For(i,n) {
For(j,n) putchar(f[i][j]);cout<<endl;
}
// }
// cout<<s<<endl;
exit(0);
gmax(ct,s)
return ;
}
int ni=i,nj=j+1;
if(j==n) ni=i+1,nj=1;
f[i][j]='U';
if(ck(i,j-1,1)&&ck(i-1,j,2,3)&&ck(i-1,j-1,3)&&ck(i-1,j+1,2)) {
dfs(ni,nj,s+1);
}
f[i][j]='L';
if(ck(i-1,j,2)&&ck(i-1,j-1,3)&&ck(i,j-1,1,3) ) {
dfs(ni,nj,s+1);
}
f[i][j]='D';
if(ck(i,j-1,3) ) {
dfs(ni,nj,s+1);
}
f[i][j]='R';
if(ck(i-1,j,3) && ck(i-1,j+1,2)) {
dfs(ni,nj,s+1);
}
f[i][j]='.';
dfs(ni,nj,s);
}
int main()
{
// freopen("D.in","r",stdin);
// freopen(".out","w",stdout);
cin>>n;
dfs(1,1);
// cout<<ct;
return 0;
}
I Julienne the Deck
给一个长度为n的排列，每次操作可以选择
#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 ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;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,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#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 Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(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)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline 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;
}
#define MAXN (101010)
int a[MAXN];
map<vector<int> ,int > h;
int main()
{
// freopen("D.in","r",stdin);
// freopen(".out","w",stdout);
ll n;cin>>n;
if(n<=2) cout<<1<<endl;else
cout<<n*2%998244353<<endl;
return 0;
// For(i,n) a[i]=i;
// For(i,1e5) {
// int p=rand()%(n-1)+2;
// reverse(a+1,a+p);
// reverse(a+p,a+n+1);
// vi t;For(i,n) t.pb(a[i]);
// h[t]=1;
// }
// cout<<SI(h)<<endl;
return 0;
}
J Knight’s Tour Redux
一个棋子每次可以从移动到
，当且仅当
和
中一个为1，一个为3.
现在你希望给出一条经过n个点的路径，要求所有横坐标恰为到
的排列，纵坐标也恰为
到
的排列.
#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 ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;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,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#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 Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(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)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline 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;
}
#define MAXN (101010)
int a[MAXN]={},n;
int st2[MAXN];
int s[4]={-1,1,-3,3};
vi st;
bool inside(int x){return 1<=x && x<=n;}
map<vi ,vi > h,h_all;
void dfs(int x,int l) {
if(l==0) {
For(i,n) {
a[i]=1;st2[l]=i;
st[l]=i;
dfs(i,1);
a[i]=0;
}
return ;
}
if(l==n) {
// cout<<"st:"<<endl;
int t=0;
For(i,n-1)if(abs(st[i])==1) t++;else t--;
if(st2[0]==1 && st2[n-1]==n){
// Rep(i,n) cout<<st2[i]<<" ";cout<<endl;
vi tv,tv2;
For(i,n-1) tv.pb(abs(st[i]));
Rep(i,n) tv2.pb(st[i]);
h[tv]=tv2;
// For(i,n-1) cout<<abs(st[i])<<' ';cout<<endl<<endl;
}
{
vi tv,tv2;
For(i,n-1) tv.pb(abs(st[i]));
Rep(i,n) tv2.pb(st[i]);
h_all[tv]=tv2;
}
return ;
}
Rep(i,4) if(inside(x+s[i]) && !a[x+s[i]]) {
a[x+s[i]]=1;
st[l]=s[i];
st2[l]=x+s[i];
dfs(x+s[i],l+1);
a[x+s[i]]=0;
}
}
int dp[112345];
int main()
{
// freopen("D.in","r",stdin);
// freopen("j.out","w",stdout);
cin>>n;
if(n==1) {
cout<<"POSSIBLE\n1 1\n";return 0;
}
if(n>=2 && n<=4) {
puts("IMPOSSIBLE");return 0;
}
if(n<=20) {
st.resize(n);
dfs(0,0);
for(auto a:h_all) {
auto p=a.fi,p2=a.fi;
Rep(i,n-1) p[i]=4-p[i];
if(h_all.count(p)) {
puts("POSSIBLE");
auto r=h_all[p],w=h_all[p2];
cout<<r[0]<<' '<<w[0]<<endl;
int x=r[0],y=w[0];
for(int i=1;i<SI(r);i++) {
x+=r[i],y+=w[i];
cout<<x<<' '<<y<<endl;
}
return 0;
cout<<"{";
for(auto j:h[p]) cout<<j<<',';cout<<"0},{";
for(auto j:h[p2]) cout<<j<<',';cout<<"0}";
return 0;
}
}
}
int st7[2][20]={{1,3,1,-3,1,3,1,0},{1,1,3,1,-3,1,3,0}};
int st9[2][20]={{1,3,3,1,-3,-3,1,3,3,0},{1,1,1,3,-1,-1,3,1,1,0}};
int st10[2][20]={{1,3,-1,-1,3,1,3,-1,-1,3,0},{1,1,3,3,-1,-3,-1,3,3,1,0}};
int st11[2][20]={{1,3,3,3,-1,-3,-3,-1,3,3,3,0},{1,1,1,1,3,-1,-1,3,1,1,1,0}};
For(i,n) dp[i]=-1;
dp[0]=0;
int bv[20]={0,6,8,9,10};
For(i,n) {
For(j,4) {
int pr=i-bv[j];
if(pr>=0 && dp[pr]!=-1) {
dp[i]=pr;
}
}
}
// For(i,n) cout<<dp[i]<<" ";cout<<endl;
puts("POSSIBLE");
int x=1,y=1;
cout<<x<<' '<<y<<endl;
n--;
while(n) {
int delta=n-dp[n];
if(delta==6) {
For(i,6) {
x+=st7[0][i];
y+=st7[1][i];
cout<<x<<" "<<y<<endl;
}
}
if(delta==8) {
For(i,8) {
x+=st9[0][i];
y+=st9[1][i];
cout<<x<<" "<<y<<endl;
}
}
if(delta==9) {
For(i,9) {
x+=st10[0][i];
y+=st10[1][i];
cout<<x<<" "<<y<<endl;
}
}
if(delta==10) {
For(i,10) {
x+=st11[0][i];
y+=st11[1][i];
cout<<x<<" "<<y<<endl;
}
}
n=dp[n];
}
return 0;
}
