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);
}
}