poj 2112 Optimal Milking 二分图多重匹配

#include<stdio.h>
#include<string.h>
#include<queue>
#include<vector>
#include<map>
#include<iostream>
using namespace std;
const int N=1024*10;
const int inf=10000000;

struct Edge
{
    int from,to,cap,flow;
};
vector<Edge>edges;
vector<int>G[N];
int s,t;
int vis[N];
int d[N];
int cur[N];

void addedge(int from,int to,int cap)
{
    Edge tp;
    tp.from=from,tp.to=to,tp.cap=cap,tp.flow=0;
    edges.push_back(tp);

    tp.from=to,tp.to=from,tp.cap=0,tp.flow=0;
    edges.push_back(tp);

    int g_size=edges.size();
    G[from].push_back(g_size-2);
    G[to].push_back(g_size-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<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<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;
        }
    }
     if(!flow) d[x] = -1;
    return flow;
}

int Maxflow()
{
    int flow=0;
    while(BFS())
    {
        memset(cur,0,sizeof(cur));
        flow+=DFS(s,inf);
    }
    return flow;
}

int gt[500][500];
int main()
{
    int l,r,mid,n,m,u,v,c,i,j,k,pt;
    int K,C,M;
    while(scanf("%d%d%d",&K,&C,&M)!=EOF)
    {
        s=0;
        n=K+C;
        t=n+1;
        memset(gt,0,sizeof(gt));
        for(i=1; i<=n; i++)
        {
            for(j=1; j<=n; j++)
            {
                scanf("%d",&gt[i][j]);
                if(gt[i][j]==0&&i!=j)
                    gt[i][j]=inf;
            }
        }
        for(k=1; k<=n; k++)
        for(i=1; i<=n; i++)
                for(j=1; j<=n; j++)
                    gt[i][j]=min(gt[i][j],gt[i][k]+gt[k][j]);
        l=0;
        r=10000;
        while(l<r)
        {
            mid=(l+r)/2;
            for(i=0; i<=t; i++) G[i].clear();
            edges.clear();
            for(i=K+1;i<=n;i++)
                addedge(s,i,1);
            for(i=1;i<=K;i++)
                addedge(i,t,M);
            for(i=K+1;i<=n;i++)
                for(j=1;j<=K;j++)
                    if(gt[i][j]<=mid&&gt[i][j]!=0)
                        addedge(i,j,1);
            int ans=Maxflow();
            if(ans>=C)
            {
                r=mid;
                pt=r;
            }
            else
                l=mid+1;
        }
        printf("%d\n",pt);
    }
    return 0;
}