hust-1024-dance party(最大流--枚举,可行流判断)

题意:

舞会上，男孩和女孩配对，求最大完全匹配个数，要求每个人最多与k个不喜欢的人配对，且每次都和不同的人配对。

分析:

将一个点拆成3个点. b,  b1, b2.   从1到n枚举ans,  判可行流.   源点s到每个b连一容量为ans边, b->b1容量inf,   b->b2容量为k,    每个g到汇点连一容量为ans的边,  g->g1容量inf,   g->g2容量为k,  如果一个boy喜欢一个girl, 则连一条边b1->g1,  容量为1,    如果一个boy讨厌一个girl,  则b2->g2, 容量为1.

满足可行流条件:   最大流==ans*n.    (n为boy或者girl数)

// File Name: 1024.cpp
// Author: Zlbing
// Created Time: 2013/9/11 18:44:02

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
using namespace std;
#define CL(x,v); memset(x,v,sizeof(x));
#define INF 0x3f3f3f3f
#define LL long long
#define REP(i,r,n) for(int i=r;i<=n;i++)
#define RREP(i,n,r) for(int i=n;i>=r;i--)
const int MAXN=;
struct Edge{
    int from,to,cap,flow;
    Edge()
    {
    }
    Edge(int from,int to,int cap,int flow):from(from),to(to),cap(cap),flow(flow)
    {
    }
};
bool cmp(const Edge& a,const Edge& b){
    return a.from < b.from || (a.from == b.from && a.to < b.to);
}
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=;i<=n;i++)G[i].clear();
        edges.clear();
    }
    void AddEdge(int from,int to,int cap){
        edges.push_back(Edge(from,to,cap,));
        edges.push_back(Edge(to,from,,));//当是无向图时，反向边容量也是cap,有向边时，反向边容量是0
        m=edges.size();
        G[from].push_back(m-);
        G[to].push_back(m-);
    }
    bool BFS(){
        CL(vis,);
        queue<int> Q;
        Q.push(s);
        d[s]=;
        vis[s]=;
        while(!Q.empty()){
            int x=Q.front();
            Q.pop();
            for(int i=;i<(int)G[x].size();i++){
                Edge& e=edges[G[x][i]];
                if(!vis[e.to]&&e.cap>e.flow){
                    vis[e.to]=;
                    d[e.to]=d[x]+;
                    Q.push(e.to);
                }
            }
        }
        return vis[t];
    }
    int DFS(int x,int a){
        if(x==t||a==)return a;
        int flow=,f;
        for(int& i=cur[x];i<(int)G[x].size();i++){
            Edge& e=edges[G[x][i]];
            if(d[x]+==d[e.to]&&(f=DFS(e.to,min(a,e.cap-e.flow)))>){
                e.flow+=f;
                edges[G[x][i]^].flow-=f;
                flow+=f;
                a-=f;
                if(a==)break;
            }
        }
        return flow;
    }
    //当所求流量大于need时就退出，降低时间
    int Maxflow(int s,int t,int need){
        this->s=s;this->t=t;
        int flow=;
        while(BFS()){
            CL(cur,);
            flow+=DFS(s,INF);
            if(flow>need)return flow;
        }
        return flow;
    }
    //最小割割边
    vector<int> Mincut(){
        BFS();
        vector<int> ans;
        for(int i=;i<edges.size();i++){
            Edge& e=edges[i];
            if(vis[e.from]&&!vis[e.to]&&e.cap>)ans.push_back(i);
        }
        return ans;
    }
    void Reduce(){
        for(int i = ; i < edges.size(); i++) edges[i].cap -= edges[i].flow;
    }
    void ClearFlow(){
        for(int i = ; i < edges.size(); i++) edges[i].flow = ;
    }
};
int n,m,k;
int s,t;
Dinic solver;
int G[][];
bool solve(int x)
{
    solver.init(t+);
    REP(i,,n)
        REP(j,,n)
        {
            if(G[i][j]==)
            {
                solver.AddEdge(*n+i,*n+j+n,);
                //printf("%d %d cap=%d\n",2*n+i,2*n+j+n,1);
            }
            else
            {
                solver.AddEdge(*n*+i,*n*+n+j,);
                //printf("%d %d cap=%d\n",2*n*2+i,2*n*2+j+n,1);
            }
        }
    REP(i,,n)
    {
        solver.AddEdge(s,i,x);
        //printf("%d %d cap=%d\n",s,i,x);
        solver.AddEdge(i,*n+i,INF);
        //printf("%d %d cap=%d\n",i,2*n+i,INF);
        solver.AddEdge(i,*n*+i,k);
        //printf("%d %d cap=%d\n",i,2*n*2+i,k);
    }
    REP(i,+n,n+n)
    {
        solver.AddEdge(i,t,x);
        //printf("%d %d cap=%d\n",i,t,x);
        solver.AddEdge(*n+i,i,INF);
        //printf("%d %d cap=%d\n",2*n+i,i,INF);
        solver.AddEdge(*n*+i,i,k);
        //printf("%d %d cap=%d\n",2*n*2+i,i,k);
    }
    int ans=solver.Maxflow(s,t,INF);
    if(ans>=n*x)
        return true;
    else return false;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&k);
         s=n**+;;
         t=s+;
        int a,b;
        CL(G,);
        REP(i,,m)
        {
            scanf("%d%d",&a,&b);
            G[a][b]=;
        }
        int ans=;
        for(int i=n;i>;i--)
        {
            //printf("Case %d:\n",i);
            if(solve(i))
            {
                ans=i;
                break;
            }
        }
        printf("%d\n",ans);
    }
    return ;
}