bzoj 4443（二分+二分图最大匹配）

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题解：

二分答案x，对于权值不大于x的点行列连边，跑二分图最大匹配，如果匹配数小于n-k+1（第k大即第n-k+1小），则left增大，否则right减小。最后分到的left+1一定是答案。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int MAXN=254,INF=0x3f3f3f3f;int n,m,K,a[MAXN][MAXN],mn=INF,mx=-INF;int mat[MAXN<<1],head[MAXN<<1],etot;bool vis[MAXN<<1];struct EDGE {int v,nxt;}e[MAXN*MAXN<<1];inline void adde(int u,int v) {e[etot].nxt=head[u],e[etot].v=v,head[u]=etot++;e[etot].nxt=head[v],e[etot].v=u,head[v]=etot++;}inline bool dfs(int p) {for (int i=head[p];~i;i=e[i].nxt) {int v=e[i].v;if (!vis[v]) {vis[v]=true;if (mat[v]==-1||dfs(mat[v])) {mat[v]=p;return true;}}}return false;}inline bool cck(int mid) {etot=0;memset(head,-1,sizeof(head));for (int i=1;i<=n;++i)for (int j=1;j<=m;++j)if (a[i][j]<=mid) adde(i,j+n);int max_match=0;memset(mat,-1,sizeof(mat));for (int i=1;i<=n;++i) {memset(vis,false,sizeof(vis));if (dfs(i)) ++max_match;}return max_match>=K;}int main() {//freopen("bzoj 4443.in","r",stdin);scanf("%d%d%d",&n,&m,&K);K=n-K+1;for (int i=1;i<=n;++i)for (int j=1;j<=m;++j)scanf("%d",&a[i][j]),mn=min(mn,a[i][j]),mx=max(mx,a[i][j]);while (mn<=mx) {int mid=(mn+mx)>>1;if (cck(mid)) mx=mid-1;else mn=mid+1;}printf("%d\n",mn);return 0;}