ACdream 1171 Matrix sum (有界费用流)

Description

sweet和zero在玩矩阵游戏，sweet画了一个N * M的矩阵，矩阵的每个格子有一个整数。zero给出N个数Ki，和M个数Kj，zero要求sweet选出一些数，满足从第 i 行至少选出了Ki个数，第j列至少选出了Kj个数。 这些数之和就是sweet要付给zero的糖果数。sweet想知道他至少要给zero多少个糖果，您能帮他做出一个最优策略吗？

Input

首行一个数T(T <= 40)，代表数据总数，接下来有T组数据。

每组数据：

第一行两个数N,M(1 <= N,M <= 50)

接下来N行，每行M个数（范围是0-10000的整数)

接下来一行有N个数Ki,表示第i行至少选Ki个元素(0 <= Ki <= M)

最后一行有M个数Kj，表示第j列至少选Kj个元素(0 <= Kj <= N)

Output

每组数据输出一行，sweet要付给zero的糖果数最少是多少

Sample Input

1
4 4
1 1 1 1
1 10 10 10
1 10 10 10
1 10 10 10
1 1 1 1
1 1 1 1

Sample Output

6

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#define inf 0x3f3f3f3f
#define LL __int64
using namespace std;
const int N = 6000;
int sum,head[N],cost,flow,cnt;
struct node{
    int to,c,w,next;
}q[N];
int mp[55][55],cur[N],f[N],dis[N],a1[N/50],a2[N/50];
int st,ed,n;
bool vis[N];
void Add(int u,int v,int cap,int cost){
    q[sum].to=v;
    q[sum].w=cap;
    q[sum].c=cost;
    q[sum].next=head[u];
    head[u]=sum++;

    q[sum].to=u;
    q[sum].w=0;
    q[sum].c=-cost;
    q[sum].next=head[v];
    head[v]=sum++;

}
int SPFA(){
    memset(vis,false,sizeof(vis));
    memset(dis,inf,sizeof(dis));
    queue<int>Q;
    while(!Q.empty())
        Q.pop();

    Q.push(st);
    cur[st] = -1;
    f[st] = inf;
    dis[st] = 0;
    while(!Q.empty()){
        int u = Q.front();
        Q.pop();
        vis[u] = false;
        for(int i = head[u];~i;i=q[i].next ){
            int v = q[i].to;
            if(dis[v] > dis[u] + q[i].c&&q[i].w ){
                dis[v]  = dis[u] + q[i].c;
                cur[v] = i;
                f[v] = min(f[u],q[i].w);
                if(!vis[v]){
                    vis[v] = true;
                    Q.push(v);
                }
            }
        }
    }

    if(dis[ed] == inf)
        return 0;
    flow += f[ed];
    cost += f[ed]*dis[ed];
    for(int i = cur[ed];~i;i = cur[q[i^1].to ]){
        q[i].w -= f[ed];
        q[i^1].w += f[ed];
    }
    return 1;
}

void init(){
    memset(head,-1,sizeof(head));
    sum = cost = flow = cnt = 0;
}

int main(){
    int m,i,j,k,cla;
    scanf("%d",&cla);
    for(int zu = 1;zu <= cla;++ zu){
        init();
        scanf("%d%d",&n,&m);
        int T = 0;
        for( i = 1;i <= n;++ i )
            for(j = 1 ;j <= m;++ j){
                scanf("%d",&mp[i][j]);
                T += mp[i][j];
            }

        for(i= 1;i <= n;++ i)
            scanf("%d",&a1[i]);
        for(i= 1;i <= n;++ i)
            scanf("%d",&a2[i]);

        st = 0;
        ed = n+m+11;
        for(i = 1;i <= n;++ i ){
            Add( st,i,m - a1[i],0 );
            Add( i,ed,m - a1[i],0 );
        }

        for(i = 1;i <= m;++ i ){
            Add( n+i,ed,n - a2[i],0 );
        }
        for(i =1 ;i <= n;++ i)
            for(j = 1;j <= m;++ j)
                Add(i,n+j,1,-mp[i][j]);
        while(SPFA());
        printf("%d\n",T + cost);
    }
    return 0;
}