目录
- 提交链接
- B . Cuckoo for Hashing [递归]
- C . Playing Fair with Cryptography
- F . Super Phyllis
- H . The Urge to Merge
- I . Xenospeak [规律模拟]
- E . Stampede! [二分 + 网络最大流(拆点)]
提交链接
https://codeforces.com/gym/100291/submit
B . Cuckoo for Hashing [递归]
int n1, n2, m;
int a1[maxn], a2[maxn];
void modify(int f,int val) {
if(f == 0) {
if(a1[val % n1] == -1) a1[val % n1] = val;
else {
int pre = a1[val % n1];
a1[val % n1] = val;
modify(1,pre);
}
} else {
if(a2[val % n2] == -1) a2[val % n2] = val;
else {
int pre = a2[val % n2];
a2[val % n2] = val;
modify(0,pre);
}
}
}
int main() {
int _ = 0;
while (~scanf("%d%d%d", &n1, &n2, &m) && m) {
memset(a1, -1, sizeof a1);
memset(a2, -1, sizeof a2);
printf("Case %d:\n", ++_);
for (int i = 1; i <= m; i++) {
modify(0,read);
}
int c1 = 0,c2 = 0;
for(int i=0;i<max(n1,n2);i++) {
if(a1[i] != -1) c1 ++;
if(a2[i] != -1) c2 ++;
}
if(c1) {
puts("Table 1");
for(int i=0; i<n1; i++) {
if(a1[i] != -1) {
printf("%d:%d\n",i,a1[i]);
}
}
}
if(c2) {
puts("Table 2");
for(int i=0; i<n2; i++) {
if(a2[i] != -1) {
printf("%d:%d\n",i,a2[i]);
}
}
}
}
return 0;
}
/**
**/
C . Playing Fair with Cryptography
队友传送门Code
#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
char a[10][10];
int b[27];
int tx[27],ty[27];
char getNext(char ch,int &idx){
if(idx==ch-'A') idx=(idx+1)%26;
if(idx==9) idx++;
ch=idx+'A';
idx=(idx+1)%26;
if(idx==9) idx++;
return ch;
}
void cul(char u,char v,char &x,char &y){
int tu=u-'A',tv=v-'A';
if(tx[tu]==tx[tv]){
//cout<<"1"<<endl;
//cout<<tu<<" "<<tx[tu]<<" "<<ty[tu]<<endl;
//cout<<tv<<" "<<tx[tv]<<" "<<ty[tv]<<endl;
//cout<<(ty[tu]+1)%5<<"***"<<(ty[tv]+1)%5<<endl;
x=a[tx[tu]][(ty[tu]+1)%5];
y=a[tx[tv]][(ty[tv]+1)%5];
//cout<<x<<" "<<y<<endl;
}else if(ty[tu]==ty[tv]){
//cout<<"2"<<endl;
x=a[(tx[tu]+1)%5][ty[tu]];
y=a[(tx[tv]+1)%5][ty[tv]];
}else{
//cout<<"3"<<endl;
x=a[tx[tu]][ty[tv]];
y=a[tx[tv]][ty[tu]];
}
}
int main(){
int _,Case=1;cin>>_;
while(_--){
string ss,s="",t="",tt;
if(Case==1) getchar();
getline(cin,ss);
getline(cin,tt);
// cout<<ss<<endl;
// cout<<tt<<endl;
for(int i=0;i<ss.size();i++){
if(ss[i]>='a'&&ss[i]<='z'){
s+=ss[i]-'a'+'A';
}
else if(ss[i]>='A'&&ss[i]<='Z'){
s+=ss[i];
}
}
for(int i=0;i<tt.size();i++){
if(tt[i]=='J') tt[i]='I';
if(tt[i]>='a'&&tt[i]<='z') t+=tt[i]-'a'+'A';
else if(tt[i]>='A'&&tt[i]<='Z') t+=tt[i];
}
//cout<<s<<endl;
//cout<<t<<endl;
int idx=0;
memset(b,0,sizeof b);
b[9]=1;
for(int i=0;i<5;i++)
for(int j=0;j<5;j++)
a[i][j]='*';
for(int i=0;i<5;i++){
for(int j=0;j<5;j++){
while(b[s[idx]-'A']&&idx<s.size()) idx++;
if(idx<s.size()){
a[i][j]=s[idx],b[s[idx]-'A']=1;
}
else break;
}
}
idx=0;
for(int i=0;i<5;i++)
for(int j=0;j<5;j++){
if(a[i][j]!='*') continue;
while(b[idx]) idx++;
a[i][j]=idx+'A';
b[idx]=1;
}
for(int i=0;i<5;i++){
for(int j=0;j<5;j++){
//cout<<a[i][j]<<" ";
tx[a[i][j]-'A']=i;
ty[a[i][j]-'A']=j;
}
//cout<<"\n";
}
idx=0;
char x,y;
string ans="";
for(int i=0;i<t.size();i+=2){
if(t[i]==t[i+1]){
cul(t[i],getNext(t[i],idx),x,y);
i--;
}
else if(i+2>t.size()){
cul(t[i],getNext(t[i],idx),x,y);
}
else cul(t[i],t[i+1],x,y);
ans=ans+x+y;
}
cout<<"Case "<<Case++<<": "<<ans<<endl;
}
return 0;
}
F . Super Phyllis
#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
int n,a[1100][1100],Case=0,idx=0;
vector<int>g[1100];
map<string,int>mp;
map<int,string>mpp;
bool vis[1100];
void init(){
idx=0;mp.clear();mpp.clear();
memset(a,0,sizeof a);
for(int i=0;i<1100;i++) g[i].clear();
}
int cul(string s){
if(!mp.count(s)) mp[s]=++idx,mpp[idx]=s;
return mp[s];
}
bool bfs(int s,int t){
queue<int>q;
memset(vis,0,sizeof vis);
vis[s]=1;q.push(s);
while(!q.empty()){
int now=q.front();q.pop();
if(now==t) return true;
for(int i=0;i<g[now].size();i++){
int j=g[now][i];
if(!a[now][j]) continue;
if(!vis[j]){
vis[j]=1;q.push(j);
}
}
}
return false;
}
int main(){
while(~scanf("%d",&n)){
if(n==0) break;
init();
for(int i=1;i<=n;i++){
string su,sv;
cin>>su>>sv;
int u=cul(su),v=cul(sv);
g[u].push_back(v);
a[u][v]=1;
}
vector<string>ans;
for(int i=1;i<=idx;i++){
for(int j=0;j<g[i].size();j++){
int now=g[i][j];
a[i][now]=0;
if(!bfs(i,now))
a[i][now]=1;
else
ans.push_back(mpp[i]+","+mpp[now]);
}
}
sort(ans.begin(),ans.end());
cout<<"Case "<<++Case<<": "<<ans.size();
for(int i=0;i<ans.size();i++)
cout<<" "<<ans[i];
cout<<"\n";
}
return 0;
}
H . The Urge to Merge
#include <bits/stdc++.h>
typedef long long ll;
#pragma region Debug
template<typename T>
std::istream& operator>> (std::istream& in, std::vector<T>& vt)
{
for (auto& i : vt) in >> i;
return in;
}
template<typename T>
std::ostream& operator<< (std::ostream& out, const std::vector<T>& vt)
{
for (auto& i : vt) out << i << " ";
out << std::endl;
return out;
}
void printAll() { }
template<typename T1, typename... T2>
void printAll(const T1& first, const T2&... second)
{
std::cout << first;
if (sizeof...(second)) {
std::cout << ", ";
}
printAll(second...);
}
template<typename T1, typename... T2>
void print(const T1& first, const T2&... second)
{
std::cout << "[";
printAll(first, second...);
std::cout << "]" << std::endl;
}
#pragma endregion
#define N (int)(5e2 + 10)
int main()
{
int n;
while (std::cin >> n && n) {
static int cas = 1;
std::vector<std::vector<int>> v(3, std::vector<int>(n)), f(n, std::vector<int>(27));
for (auto& str : v) {
for (auto& ch : str) {
std::cin >> ch;
}
}
/*
f[i][j] : 1 - i 列所有的数据，最后一列状态为j的最大值
f[i][j] = max(f[i - 1][k] + val(j))
*/
auto judge = [](const std::vector<int>& tmp) {
for (int i = 0; i < tmp.size() - 1; i++) {
if (tmp[i] == 1) {
if (tmp[i + 1] != 1) {
return false;
}
i += 1;
}
}
return true;
};
std::vector<std::vector<int>> allSta;
std::vector<int> allNum;
for (int i = 0; i < 27; i++) {
std::vector<int> tmp;
int num = i;
while (num) {
tmp.push_back(num % 3);
num /= 3;
}
while (tmp.size() != 3) tmp.push_back(0);
std::reverse(tmp.begin(), tmp.end());
if (judge(tmp)) {
allSta.push_back(tmp);
allNum.push_back(i);
}
}
auto calc1 = [&](const std::vector<int>& sta, int col)
{
int res = 0;
for (int i = 0; i < sta.size() - 1; i++) {
if (sta[i] == 1) {
res += v[i][col] * v[i + 1][col];
i += 1;
}
}
return res;
};
//print(calc1({ 0,1,1 }, 0));
for (auto& tmp : allSta) {
//print(tmp[0], tmp[1], tmp[2]);
}
const int& si = allNum.size();
for (int i = 0; i <= 0; i++) {
for (int j = 0; j < si; j++) {
auto& sta = allSta[j];
auto& num = allNum[j];
f[i][num] = calc1(sta, i);
}
}
auto judge2 = [](const std::vector<int>& curSta, const std::vector<int>& preSta) {
for (int i = 0; i < curSta.size(); i++) {
if (curSta[i] == 2) {
if (preSta[i] == 1) {
return false;
}
if (preSta[i] == 2) {
return false;
}
}
}
return true;
};
auto calc2 = [&](const std::vector<int>& curSta, const std::vector<int>& preSta, int col)
{
int res = 0;
for (int i = 0; i < curSta.size() - 1; i++) {
if (curSta[i] == 1) {
res += v[i][col] * v[i + 1][col];
i += 1;
}
else if (curSta[i] == 2) {
res += v[i][col] * v[i][col - 1];
}
}
if (curSta[curSta.size() - 1] == 2) {
res += v[curSta.size() - 1][col] * v[curSta.size() - 1][col - 1];
}
return res;
};
/*
0 : 先不合并
1 : 上下合并
2 : 突出
*/
int mx = 0;
for (int i = 1; i < n; i++) {
for (int j = 0; j < si; j++) {
auto& curSta = allSta[j];
auto& curNum = allNum[j];
for (int k = 0; k < si; k++) {
auto& preSta = allSta[k];
auto& preNum = allNum[k];
if (judge2(curSta, preSta)) {
f[i][curNum] = std::max(f[i][curNum], f[i - 1][preNum] + calc2(curSta, preSta, i));
}
}
if (i == n - 1) {
mx = std::max(mx, f[i][curNum]);
}
}
}
printf("Case %d: %d\n", cas++, mx);
//std::cout << f[0][4] << std::endl;
}
return 0;
}
I . Xenospeak [规律模拟]
其中，num[i]表示长度为i的字符串有多少个
然后确定了字符串的长度进行构造
如果说前一个为a开头的，那么说后面可以拼接任意的字符串，转移到len-1的情况
如果说是b开头的，只能是bb并且转移到len-2的情况
根据规律进行模拟
ll num[1007];
string get(ll a) {
int len = 1;
while (num[len] < a) a -= num[len],len++;
string s;
s.clear();
int fa = 0;
for (int i = len; i >= 1; i--) {
ll need = num[i-1];/// 看两位，注意没有两位的情况，比如当前i为1
if(i > 1) need += num[i-2];
if(need >= a) s.push_back('a'),fa = 1;
else {
if(fa == 0) {
s.push_back('b');
s.push_back('b');
a -= need;
-- i;/// 加了两个需要再减去一个长度
}else {
s.push_back('b');
a -= need;
}
}
}
return s;
}
int main() {
num[0] = num[1] = 1;
for (int i = 2; i <= 1000; i++) num[i] = num[i - 1] + num[i - 2] + num[i - 2];
ll n, m;
int _ = 0;
while (cin >> n >> m && (n || m)) {
printf("Case %d: %s %s\n", ++_, get(n * (m - 1) + 1).c_str(), get(n * m).c_str());
}
return 0;
}
/**
**/
E . Stampede! [二分 + 网络最大流(拆点)]
没有技巧，全是感情
可以参考：博客1 博客2提交链接 Code:
/*** keep hungry and calm PushyTao!***/
#pragma GCC optimize(2)
#pragma G++ optimize(2)
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll inf = 9223372036854775807;
const ll INF = 0x3f3f3f3f;
const int maxn = 1e6 + 7000;
struct Edge {
int u, v;
ll cap, flow;
Edge(int _u, int _v, ll _cap, ll _flow) {
u = _u, v = _v;
cap = _cap, flow = _flow;
}
};
struct Dinic {
vector<Edge> edge;
vector<int> G[maxn << 1];
ll dis[maxn << 1], cur[maxn << 1];
int n, s, t;
bool vis[maxn << 1];
void init(int x, int _s, int _t) {
n = x;
for (int i = 0; i <= n; i++) G[i].clear();
s = _s, t = _t;
edge.clear();
}
void add(int u, int v, ll cap) {
edge.push_back(Edge(u, v, cap, 0));
edge.push_back(Edge(v, u, 0, 0));
G[u].push_back(edge.size() - 2);
G[v].push_back(edge.size() - 1);
}
bool bfs(int s, int t) {
queue<int> que;
memset(vis, 0, sizeof vis);
// memset(dis,0,sizeof dis);
dis[s] = 0;
que.push(s);
vis[s] = 1;
while (que.size()) {
int u = que.front();
que.pop();
for (int i = 0; i < (int)G[u].size(); i++) {
int id = G[u][i];
int to = edge[id].v;
if (!vis[to] && edge[id].cap > edge[id].flow) {
dis[to] = dis[u] + 1;
que.push(to);
vis[to] = 1;
}
}
}
return vis[t];
}
ll dfs(int s, int t, ll rest) {
if (s == t || rest == 0) return rest;
ll Flow = 0, f;
for (ll &i = cur[s]; i < G[s].size(); i++) {
Edge &e = edge[G[s][i]];
if (dis[s] + 1 == dis[e.v] && (f = dfs(e.v, t, min(rest, e.cap - e.flow))) > 0) {
e.flow += f;
edge[G[s][i] ^ 1].flow -= f;
Flow += f;
rest -= f;
if (rest == 0) break;
}
}
return Flow;
}
ll getMaxFlow(int s, int t) {
ll ans = 0;
while (bfs(s, t)) {
memset(cur, 0, sizeof cur);
ans += dfs(s, t, 0x3f3f3f3f);
}
return ans;
}
} solve;
int n;
string s[maxn];
int getId(int x, int y, int f) {
return x * n + y + 1 + f * n * n;
}
int dx[4] = {0, 0, -1, 1};
int dy[4] = {-1, 1, 0, 0};
bool in(int x, int y) {
if (x >= 0 && x <= n - 1 && y >= 0 && y <= n - 1 && s[x][y] == '.') return true;
return false;
}
bool check(int v) {
int st = 0, ed = n * n * (v + 1) + 1;
solve.init(1000001, st, ed);
for (int i = 0; i < n; i++) {
solve.add(st, getId(i, 0, 0), 1);
solve.add(getId(i, n - 1, v), ed, 1);
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (s[i][j] == '.') {
for (int k = 0; k < v; k++) {
solve.add(getId(i, j, k), getId(i, j, k + 1), 1);
for (int t = 0; t < 4; t++) {
int tox = i + dx[t];
int toy = j + dy[t];
if (in(tox, toy)) {
solve.add(getId(i, j, k), getId(tox, toy, k + 1), 1);
}
}
}
}
}
}
return solve.getMaxFlow(st, ed) == n;
}
int main() {
int _ = 0;
while (cin >> n && n) {
for (int i = 0; i < n; i++) cin >> s[i];
int l = 0, r = n * n + 10;
int ans = 0;
while (l < r) {
int mid = ((l + r) >> 1);
if (check(mid)) ans = mid, r = mid;
else l = mid + 1;
}
printf("Case %d: %d\n", ++_, ans);
}
return 0;
}
