poj 3189 Steady Cow Assignment 二分图多重匹配

#include<stdio.h>
#include<string.h>
#include<queue>
#include<vector>
using namespace std;
const int N=1200;
const int inf=1<<24;

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];
int g[N][30];
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 st,int ed)
{
    int flow=0;
    while(BFS())
    {
        memset(cur,0,sizeof(cur));
        flow+=DFS(st,inf);
    }
    return flow;
}

int main()
{
    int i,j,k,n,m,u,v,c,mp[N],r,l,f;
    while(~scanf("%d%d",&n,&m))
    {

        s=0;
        t=n+m+1;
        for(i=1; i<=n; i++)
            for(j=1; j<=m; j++)
                scanf("%d",&g[i][j]);
        for(i=1; i<=m; i++)
            scanf("%d",&mp[i]);

        l=0,r=m;
        while(l<r)
        {
            f=0;
            k=(l+r)/2;
            for(int x=1; x+k<=m; x++)
            {
                //k=(l+r)/2;
                edges.clear();
                for(i=0; i<N; i++) G[i].clear();
                for(i=1; i<=n; i++)
                    AddEdge(s,i,1);
                for(i=n+1; i<=n+m; i++)
                    AddEdge(i,t,mp[i-n]);
                for(i=1; i<=n; i++)
                    for(j=x; j<=x+k; j++)
                        AddEdge(i,g[i][j]+n,1);
                if(Maxflow(s,t)==n)
                {
                    f=1;
                    break;
                }
            }
            if(f==1) r=k;
            else l=k+1;
        }
        printf("%d\n",r+1);
    }
    return 0;
}