【链接】 我是链接,点我呀:)
【题意】

在这里输入题意

【题解】

男生和女生每个人都分身成两个节点
即x[1],x[2]和y[1],y[2]

然后如果i和j不相互喜欢
那么add(x[i][2],y[j][2],1)
如果相互喜欢的话
add(x[i][1],y[j][1],1)

然后对于每个男生
add(x[i][1],y[i][1],k)

对于每个女生
add(y[i][2],y[i][1],k)

然后对于每个男生
add(s,x[i][1],mid)
add(y[i][1],t,mid)

这里的mid是二分的值。

这个mid就是舞会的轮数了。

如果能够满流显然每个人都能配对跳mid支舞
显然是有单调性的。
(然后发现原来的dicnic的模板是错的。。边数的计算要特别careful.

【代码】

#include <bits/stdc++.h>
using namespace std;const int N = 50;struct abc{int nex,en,flow;
}bian[2][N*N*2+8*N+10];int n,k,x[N+10][3],y[N+10][3],fir[2][N*4+10],tfir[N*4+10],totm,deep[N*4+10];
int cnt = 0;
bool bo[N+10][N+10];
char s[N+10];
queue<int> dl;//每个男人分为(x1,x2)
//每个女人分为(y1,y2)void add(int x,int y,int cost,int i){bian[i][totm].nex = fir[i][x];fir[i][x] = totm;bian[i][totm].en = y,bian[i][totm].flow = cost;totm++;bian[i][totm].nex = fir[i][y];fir[i][y] = totm;bian[i][totm].en = x,bian[i][totm].flow = 0;totm++;
}bool bfs(int s,int t){dl.push(s);memset(deep,255,sizeof deep);deep[s] = 0;while (!dl.empty()){int x = dl.front();dl.pop();for (int temp = fir[1][x]; temp!= -1 ;temp = bian[1][temp].nex){int y = bian[1][temp].en;if (deep[y]==-1 && bian[1][temp].flow>0){deep[y] = deep[x] + 1;dl.push(y);}}}return deep[t]!=-1;
}int dfs(int x,int t,int limit){if (x == t) return limit;if (limit == 0) return 0;int cur,f = 0;for (int temp = tfir[x];temp!=-1;temp = bian[1][temp].nex){tfir[x] = temp;int y = bian[1][temp].en;if (deep[y] == deep[x] + 1 && (cur = dfs(y,t,min(limit,bian[1][temp].flow))) ){f += cur;limit -= cur;bian[1][temp].flow -= cur;bian[1][temp^1].flow += cur;if (!limit) break;}}return f;
}int get_flow(){int now = 0;while (bfs(0,cnt)){for (int i = 0;i <= cnt;i++) tfir[i] = fir[1][i];int xxx = dfs(0,cnt,10000);now+=xxx;}return now;
}int main(){//freopen("F:\\program\\rush\\rush_in.txt","r",stdin);ios::sync_with_stdio(0),cin.tie(0);memset(fir[0],255,sizeof fir[0]);cin >> n >> k;for (int i = 1;i <= n;i++){cin >> (s+1);for (int j = 1;j <= n;j++)if (s[j]=='Y'){bo[i][j] = 1;}}for (int i = 1;i <= n;i++){for (int j = 1;j <= 2;j++)x[i][j] = ++cnt;}for (int i = 1;i <= n;i++)for (int j = 1;j <= 2;j++)y[i][j] = ++cnt;cnt++;for (int i = 1;i <= n;i++){for (int j = 1;j <= n;j++){if (bo[i][j])add(x[i][1],y[j][1],1,0);else{add(x[i][2],y[j][2],1,0);}}}for (int i = 1;i <= n;i++){add(x[i][1],x[i][2],k,0);add(y[i][2],y[i][1],k,0);}int l = 0,r = n+1,temp1 = 0;while (l<=r){int i = (l+r)>>1;for (int j = 0;j < totm;j++) bian[1][j] = bian[0][j];for (int j = 0;j <= cnt;j++) fir[1][j] = fir[0][j];int tt = totm;for (int j = 1;j <= n;j++){add(0,x[j][1],i,1);add(y[j][1],cnt,i,1);}int temp = get_flow();if (temp!=i*n){r = i-1;}else {temp1 = i;l = i+1;}totm = tt;}cout<<temp1<<endl;return 0;
}