uva 11082 (网络流 +Dinic )

根据 矩阵和二分图的联系 建图 ，矩阵的每一行对应着一个节点，矩阵的每一列对应着一个节点。

对于源点s到行的点-1 是因为最大流中有可能会出现0流。

代码： 

#include<bits/stdc++.h>

using namespace std;
const int inf=0x3f3f3f3f;
const int N=1000;
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[N];
    bool vis[N];
    int d[N];
    int cur[N];

    void init()
    {
        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));
        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;
        for (int i=0; i<N; i++) d[i] = inf;
        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;
            }
        }
        return flow;
    }

    int maxflow(int s,int t,int R,int C)
    {
        this->s=s; this->t=t;
        int flow=0;
        while(bfs()){
            memset(cur,0,sizeof(cur));
            flow+=dfs(s,inf);
        }

        //printf("%d %d\n",oka(),okb(R,C));

       for (int i=1; i<=R; i++)
            {
                int j;
                for (j=1; j<G[i].size()-1; j++)
                    cout<<edges[G[i][j]].flow+1<<' ';
                cout<<edges[G[i][j]].flow+1<<endl;
            }

        return flow;
    }

}a;


int R,C;
int na[N];
int nb[N];

int main()
{
    int T;
    int kk=0,ff=0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d",&R,&C);
        for(int i=1;i<=R;i++){
            scanf("%d",&na[i]);
        }
        for(int i=R;i>=1;i--){
            na[i]=na[i]-na[i-1]-C;
        }
        for(int i=1;i<=C;i++){
            scanf("%d",&nb[i]);
        }
        for(int i=C;i>=1;i--){
            nb[i]=nb[i]-nb[i-1]-R;
        }
        a.init();
        a.s=0;  a.t=R+C+1;
        for(int i=1;i<=R;i++)
        {
            a.addedge(0,i,na[i]);
        }
        for(int i=1;i<=C;i++){
            a.addedge(R+i,R+C+1,nb[i]);
        }
        for(int i=1;i<=R;i++){
            for(int j=1;j<=C;j++){
                a.addedge(i,R+j,19);
            }
        }

        printf("Matrix %d\n",++kk);
        a.maxflow(0,R+C+1,R,C);
        printf("\n");
    }
    return 0;
}