add rom some som max rds initial else RoCE

Mirko works on a pig farm that consists of M locked pig-houses and Mirko can‘t unlock any pighouse because he doesn‘t have the keys. Customers come to the farm one after another. Each of them has keys to some pig-houses and wants to buy a certain number of pigs.
All data concerning customers planning to visit the farm on that particular day are available to Mirko early in the morning so that he can make a sales-plan in order to maximize the number of pigs sold.
More precisely, the procedure is as following: the customer arrives, opens all pig-houses to which he has the key, Mirko sells a certain number of pigs from all the unlocked pig-houses to him, and, if Mirko wants, he can redistribute the remaining pigs across the unlocked pig-houses.
An unlimited number of pigs can be placed in every pig-house.
Write a program that will find the maximum number of pigs that he can sell on that day. Input The first line of input contains two integers M and N, 1 <= M <= 1000, 1 <= N <= 100, number of pighouses and number of customers. Pig houses are numbered from 1 to M and customers are numbered from 1 to N.
The next line contains M integeres, for each pig-house initial number of pigs. The number of pigs in each pig-house is greater or equal to 0 and less or equal to 1000.
The next N lines contains records about the customers in the following form ( record about the i-th customer is written in the (i+2)-th line):
A K1 K2 ... KA B It means that this customer has key to the pig-houses marked with the numbers K1, K2, ..., KA (sorted nondecreasingly ) and that he wants to buy B pigs. Numbers A and B can be equal to 0. Output The first and only line of the output should contain the number of sold pigs. Sample Input

3 3
3 1 10
2 1 2 2
2 1 3 3
1 2 6

Sample Output

7

題意 ： 有N個顧客，有M個豬圈，每個豬圈有一定的豬，在開始的時候豬圈都是關閉的，

顧客來買豬，顧客打開某個豬圈，可以在其中挑選一定的豬的數量，在這個顧客走後，可以在打開的

豬圈中將某個豬圈的一些豬牽到另外一個打開的豬圈，然後所有的豬圈會關閉，這樣下一個顧客來了繼續上面的工作

思路分析 ：
　　參考此文章即可 ：http://wenku.baidu.com/view/0ad00abec77da26925c5b01c.html
代碼示例 ：
　　

const int maxn = 1e4+5;
const int inf = 0x3f3f3f3f;

int m, n;
int a[1005];
vector<int>ve[1005];
struct node
{
    int to, flow;
    int next;
}e[maxn];
int head[maxn];
int cnt = 0;
int s, t;

void addedge(int u, int v, int w){
    e[cnt].next = head[u], e[cnt].to = v, e[cnt].flow = w, head[u] = cnt++;
    e[cnt].next = head[v], e[cnt].to = u, e[cnt].flow = 0, head[v] = cnt++;
}
int belong[1005];
int dep[1005], que[maxn];

bool bfs(int s, int t){
    int head1 = 0, tail = 1;
    memset(dep, 0, sizeof(dep));
    que[0] = s; dep[s] = 1;
    while(head1 < tail){
        int v = que[head1++];
        for(int i = head[v]; i != -1; i = e[i].next){
            int to = e[i].to;
            if (e[i].flow && !dep[to]) {
                dep[to] = dep[v]+1;
                que[tail++] = to;
            }
        }
    }
    return dep[t];
}
int dfs(int u, int f1){
    if (f1 == 0 || u == t) return f1;
    
    int f = 0;
    for(int i = head[u]; i != -1; i = e[i].next){
        int to = e[i].to;
        if (e[i].flow && dep[to] == dep[u]+1){
            int x = dfs(to, min(e[i].flow, f1));
            e[i].flow -= x; e[i^1].flow += x;
            f1 -= x, f += x;
            if (f1 == 0) return f;
        }
    }
    if (!f) dep[u] = -2;
    return f;
}

int maxflow(int s, int t){
    int ans = 0;
    while(bfs(s, t)){
        ans += dfs(s, inf); 
    }
    return ans;
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int x, y;
    cin >> m >> n;
    s = 0, t = n+1;
    memset(head, -1, sizeof(head));
    for(int i = 1; i <= m; i++) scanf("%d", &a[i]);
    for(int i = 1; i <= n; i++){
        scanf("%d", &x);
        for(int j = 1; j <= x; j++) {
            scanf("%d", &y);
            ve[i].push_back(y);
        }
        scanf("%d", &y);
        addedge(i, t, y);
    }
    for(int i = 1; i <= n; i++){
        for(int j = 0; j < ve[i].size(); j++){
            int v = ve[i][j]; 
            if (!belong[v]) {
                belong[v] = i;
                addedge(s, i, a[v]);
            }
            else {
                addedge(belong[v], i, inf);
                belong[v] = i;
            }
        }
    } 
    printf("%d\n", maxflow(s, t));
    return 0;
}

小豬分配 ， 最大流