图论--网络流--最大流 POJ 2289 Jamie's Contact Groups （二分+限流建图）

3 2
John 0 1
Rose 1
Mary 1
5 4
ACM 1 2 3
ICPC 0 1
Asian 0 2 3
Regional 1 2
ShangHai 0 2
0 0

2
2

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#define INF 1e9
using namespace std;
const int maxn=1500+5;
 
struct Edge
{
    int from,to,cap,flow;
    Edge(){}
    Edge(int f,int t,int c,int fl):from(f),to(t),cap(c),flow(fl){}
};
 
struct Dinic
{
    int n,m,s,t;
    vector<Edge> edges;
    vector<int> G[maxn];
    int d[maxn];
    int cur[maxn];
    bool vis[maxn];
 
    void init(int n,int s,int t)
    {
        this->n=n, this->s=s, this->t=t;
        edges.clear();
        for(int i=0;i<n;i++) G[i].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) );
        m = edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }
 
    bool BFS()
    {
        queue<int> Q;
        memset(vis,0,sizeof(vis));
        vis[s]=true;
        d[s]=0;
        Q.push(s);
        while(!Q.empty())
        {
            int x=Q.front(); Q.pop();
            for(int i=0;i<G[x].size();++i)
            {
                Edge& e=edges[G[x][i]];
                if(!vis[e.to] && e.cap>e.flow)
                {
                    vis[e.to]=true;
                    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<G[x].size();++i)
        {
            Edge& e=edges[G[x][i]];
            if(d[e.to]==d[x]+1 && (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 max_flow()
    {
        int ans=0;
        while(BFS())
        {
            memset(cur,0,sizeof(cur));
            ans +=DFS(s,INF);
        }
        return ans;
    }
}DC;
 
int n,m;
vector<int> g[maxn];//g[i]中保存第i个人可被分到的组编号
bool solve(int limit)
{
    int src=0, dst=n+m+1;
    DC.init(2+n+m,src,dst);
    for(int i=1;i<=n;i++) DC.AddEdge(src,i,1);
    for(int i=1;i<=m;i++) DC.AddEdge(n+i,dst,limit);
    for(int i=1;i<=n;i++)
    for(int j=0;j<g[i].size();++j)
        DC.AddEdge(i,g[i][j],1);
    return DC.max_flow() == n;
}
 
int main()
{
    while(scanf("%d%d",&n,&m)==2)
    {
        if(n==0 && m==0) break;
        for(int i=1;i<=n;i++) g[i].clear();
        for(int i=1;i<=n;i++)
        {
            char str[100];
            scanf("%s",str);
            while(1)
            {
                int x;
                scanf("%d",&x);
                g[i].push_back(x+1+n);//注意这里压入的已经是处理后的编号了
                char ch=getchar();
                if(ch=='\n') break;
            }
        }
        int L=0,R=n;
        while(R>L)
        {
            int mid=L+(R-L)/2;
            if(solve(mid)) R=mid;
            else L=mid+1;
        }
        printf("%d\n",R);
    }
    return 0;
}