acdream--Matrix sum

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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

14 41 1 1 11 10 10 101 10 10 101 10 10 101 1 1 11 1 1 1

Sample Output

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 0x3f3f3f3fint 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);    }}

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