acdream--Matrix sum

Problem 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

思路：超源向行连边，容量为m-Ki，费用为0.列向超汇连边，容量为n-Kj，费用为0.行列连容量为1，费用为-key的边。当spfa找到的路径cost>=0就可停止。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define maxn 3800
#define maxm 48000
#define inf 0x3f3f3f3f
int first[maxn];
int key[108][108];
int vv[maxm],ww[maxm],nxt[maxm],cst[maxm];
int e;
int pre[maxn],pos[maxn];
int dis[maxn],que[maxn*10];
bool vis[maxn];
inline int min(int a,int b)
{
    return a > b?b:a;
}
void addEdge(int u,int v,int w,int cost)
{
    vv[e] = v;
    ww[e] = w;
    cst[e] = cost;
    nxt[e] = first[u];
    first[u] = e++;
    vv[e] = u;
    ww[e] = 0;
    cst[e] = -cost;
    nxt[e] = first[v];
    first[v] = e++;
}
 
int spfa(int s,int t)
{
    memset(pre,-1,sizeof(pre));
    memset(vis,0,sizeof(vis));
    int head,tail;
    head = tail = 0;
    for(int i = 0;i < maxn;i++)
        dis[i] = inf;
    que[tail++] = s;
    pre[s] = s;
    dis[s] = 0;
    vis[s] = 1;
    while(head < tail)
    {
        int u = que[head++];
        vis[u] = 0;
        for(int i = first[u];i != -1;i = nxt[i])
        {
            int v = vv[i];
            if(ww[i] > 0 && dis[u] + cst[i] < dis[v])
            {
                dis[v] = dis[u] + cst[i];
                pre[v] = u;
                pos[v] = i;
                if(!vis[v])
                {
                    vis[v] = 1;
                    que[tail++] = v;
                }
            }
        }
    }
    return pre[t] != -1;
}
 
int MinCostFlow(int s,int t,int flow)
{
    int cost = 0;
    int nowflow = 0;
    while(spfa(s,t))
    {
        int f = inf;
        for(int i = t;i != s;i = pre[i])
            if(ww[pos[i]] < f)   f = ww[pos[i]];
        if(dis[t] >= 0)  break;
        f = min(flow - nowflow,f);
        nowflow += f;   cost += dis[t]*f;
        for(int i = t;i != s;i = pre[i])
        {
            ww[pos[i]] -= f;
            ww[pos[i]^1] += f;
        }
        if(nowflow == flow) break;
    }
    return cost;
}

int main()
{
    //freopen("in.txt","r",stdin);
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        e = 0;
        memset(first,-1,sizeof(first));
        int sum = 0;
        for(int i = 1;i <= n;i++)
        {
            for(int j = 1;j <= m;j++)
            {
                int a;  scanf("%d",&a);
                sum += a;
                addEdge(i,n+j,1,-a);
            }
        }
        for(int i = 1;i <= n;i++)
        {
            int a;  scanf("%d",&a);
            addEdge(0,i,m-a,0);
        }
        for(int i = 1;i <= m;i++)
        {
            int a;  scanf("%d",&a);
            addEdge(n+i,n+m+1,n-a,0);
        }
        int fuck = MinCostFlow(0,n+m+1,inf);
        printf("%d\n",sum+fuck);
    }
}