POJ 2289 最大流

//二分最大值mid，源点连向n个联系人权值为1，联系人连向它可以属于的分类，m个分类与汇点建边权值为mid（该分类最多mid个人）
//看最大流是否等于n。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int maxn=2009;
const int inf=0x7fffffff;
vector<int>v[maxn];
struct Edge{
    int from,to,cap,flow;
    Edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f){}
};
struct Dinic{
    int n,m,s,t;
    vector<Edge>edges;
    vector<int>g[maxn];
    bool vis[maxn];
    int d[maxn];
    int cur[maxn];
    void Init(int n){
        this->n=n;
        for(int i=0;i<n;i++) g[i].clear();
        edges.clear();
    }
    void Addedge(int from,int to,int cap){
        edges.push_back(Edge(from,to,cap,0));
        edges.push_back(Edge(to,from,0,0));//反向弧
        m=edges.size();
        g[from].push_back(m-2);
        g[to].push_back(m-1);
    }
    bool Bfs(){
        memset(vis,0,sizeof(vis));
        queue<int>q;
        q.push(s);
        d[s]=0;
        vis[s]=1;
        while(!q.empty()){
            int x=q.front();q.pop();
            for(int i=0;i<(int)g[x].size();i++){
                Edge &e=edges[g[x][i]];
                if(!vis[e.to]&&e.cap>e.flow){
                    vis[e.to]=1;
                    d[e.to]=d[x]+1;
                    q.push(e.to);
                }
            }
        }
        return vis[t];
    }
    int Dfs(int x,int a){
        if(x==t||a==0) return a;
        int flow=0,f;
        for(int&i=cur[x];i<(int)g[x].size();i++){
            Edge &e=edges[g[x][i]];
            if(d[x]+1==d[e.to]&&(f=Dfs(e.to,min(a,e.cap-e.flow)))>0){
                e.flow+=f;
                edges[g[x][i]^1].flow-=f;
                flow+=f;
                a-=f;
                if(a==0) break;
            }
        }
        return flow;
    }
    int Maxflow(int s,int t){
        this->s=s;this->t=t;
        int flow=0;
        while(Bfs()){
            memset(cur,0,sizeof(cur));
            flow+=Dfs(s,inf);
        }
        return flow;
    }
}dc;

int n,m;
char ch[20];

bool solve(int mid){
    dc.Init(n+m+2);
    for(int i=1;i<=n;i++){
        dc.Addedge(0,i,1);
        for(int j=0;j<(int)v[i].size();j++)
            dc.Addedge(i,v[i][j]+n,1);
    }
    for(int i=n+1;i<=n+m;i++){
        dc.Addedge(i,n+m+1,mid);
    }
    return dc.Maxflow(0,n+m+1)==n;
}

int main()
{
    while(scanf("%d%d",&n,&m)&&(n+m)){
        for(int i=1;i<=n;i++){
            scanf("%s",ch);
            while(getchar()!=‘\n‘){
                int x;
                scanf("%d",&x);
                x++;
                v[i].push_back(x);
            }
        }
        int l=0,r=n,mid,ans=0;
        while(l<=r){
            mid=(l+r)/2;
            if(solve(mid)){
                ans=mid;
                r=mid-1;
            }
            else l=mid+1;
        }
        printf("%d\n",ans);
        for(int i=1;i<=n;i++) v[i].clear();
    }
    return 0;
}