POJ1274: The Perfect Stall 题解

这是一个二分图匹配的模板题

直接贴代码吧

#include <cstdio>#include <iostream>#include <cstring>#include <string>#include <cmath>#include <algorithm>#include <cstdlib>#include <utility>#include <map>#include <stack>#include <set>#include <vector>#include <queue>#include <deque>#define x first#define y second#define mp make_pair#define pb push_back#define LL long long#define Pair pair<int,int>#define LOWBIT(x) x & (-x)using namespace std;const int MOD=1e9+7;const int INF=0x7ffffff;const int magic=348;int depth[100048];queue<int> q;int t,tot,head[100048],nxt[200048],to[200048],f[200048];inline void addedge(int s,int t,int cap){to[++tot]=t;nxt[tot]=head[s];head[s]=tot;f[tot]=cap;to[++tot]=s;nxt[tot]=head[t];head[t]=tot;f[tot]=0;}int n,m;bool bfs(){int i,x,y;for (i=0;i<=t;i++) depth[i]=-1;depth[0]=0;q.push(0);while (!q.empty()){x=q.front();q.pop();for (i=head[x];i;i=nxt[i]){y=to[i];if (f[i] && depth[y]==-1){depth[y]=depth[x]+1;q.push(y);}}}if (depth[t]==-1) return false; else return true;}int dfs(int x,int maxf){if (x==t) return maxf;int i,y,now,minf,ans=0;for (i=head[x];i;i=nxt[i]){y=to[i];if (f[i] && depth[y]==depth[x]+1){minf=min(maxf-ans,f[i]);now=dfs(y,minf);f[i]-=now;f[i^1]+=now;ans+=now;}}return ans;}int main (){int i,j,x,ax;while (scanf("%d%d",&n,&m)!=EOF){t=n+m+1;tot=1;for (i=0;i<=t;i++) head[i]=0;for (i=1;i<=n;i++) addedge(0,i,1);for (i=1;i<=m;i++) addedge(n+i,t,1);for (i=1;i<=n;i++){scanf("%d",&ax);for (j=1;j<=ax;j++){scanf("%d",&x);addedge(i,n+x,1);}}int ans=0;while (bfs()) ans+=dfs(0,2e9);printf("%d\n",ans);}return 0;}