# 小猪分配 ， 最大流

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题意 ： 有N个顾客，有M个猪圈，每个猪圈有一定的猪，在开始的时候猪圈都是关闭的，

```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 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;
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;
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);
}
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;
}
else {
belong[v] = i;
}
}
}
printf("%d\n", maxflow(s, t));
return 0;
}
```

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