POJ1149(最大流)

PIGS

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 21678		Accepted: 9911

Description

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

Source

Croatia OI 2002 Final Exam - First day

 

 

 

 

//2017-08-23
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>

using namespace std;

const int N = 510;
const int M = 1010;
const int INF = 0x3f3f3f3f;
int head[N], tot;
struct Edge{
    int next, to, w;
}edge[N<<4];

void add_edge(int u, int v, int w){
    edge[tot].w = w;
    edge[tot].to = v;
    edge[tot].next = head[u];
    head[u] = tot++;

    edge[tot].w = 0;
    edge[tot].to = u;
    edge[tot].next = head[v];
    head[v] = tot++;
}

struct Dinic{
    int level[N], S, T;
    void init(int _S, int _T){
        S = _S;
        T = _T;
        tot = 0;
        memset(head, -1, sizeof(head));
    }
    bool bfs(){
        queue<int> que;
        memset(level, -1, sizeof(level));
        level[S] = 0;
        que.push(S);
        while(!que.empty()){
            int u = que.front();
            que.pop();
            for(int i = head[u]; i != -1; i = edge[i].next){
                int v = edge[i].to;
                int w = edge[i].w;
                if(level[v] == -1 && w > 0){
                    level[v] = level[u]+1;
                    que.push(v);
                }
            }
        }
        return level[T] != -1;
    }
    int dfs(int u, int flow){
        if(u == T)return flow;
        int ans = 0, fw;
        for(int i = head[u]; i != -1; i = edge[i].next){
            int v = edge[i].to, w = edge[i].w;
            if(!w || level[v] != level[u]+1)
              continue;
            fw = dfs(v, min(flow-ans, w));
            ans += fw;
            edge[i].w -= fw;
            edge[i^1].w += fw;
            if(ans == flow)return ans;
        }
        if(ans == 0)level[u] = 0;
        return ans;
    }
    int maxflow(){
        int flow = 0;
        while(bfs())
          flow += dfs(S, INF);
        return flow;    
    }

}dinic;

int house[M];

int main()
{
    std::ios::sync_with_stdio(false);
    //freopen("input.txt", "r", stdin);
    int n, m;
    while(cin>>m>>n){
        int s = 0, t = n+1;
        dinic.init(s, t);
        for(int i = 1; i <= m; i++)
          cin>>house[i];
        int k, v;
        int book[M];
        memset(book, 0, sizeof(book));
        for(int i = 1; i <= n; i++){
            cin>>k;
            int weight = 0;
            while(k--){
                cin>>v;
                if(!book[v]){
                    book[v] = i;
                    weight += house[v];
                }else{
                    add_edge(book[v], i, INF);
                }
            }
            if(weight)add_edge(s, i, weight);
            cin>>v;
            add_edge(i, t, v);
        }
        cout<<dinic.maxflow()<<endl;
    }
    return 0;

}