POJ 1149 PIGS

PIGS

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 20892		Accepted: 9549

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

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<vector>
#define maxn 1050
using namespace std;
int dep[110],m,n,pig[maxn],house[maxn];
int vis[110],map[110][110];
#define INF 0x7fffffff
bool BFS(){
    memset(dep,-1,sizeof(dep));
    queue<int>q;
    dep[0]=0;q.push(0);
    while(!q.empty()){
        int u=q.front();q.pop();
        for(int v=1;v<=n+1;v++){
            if(map[u][v]>0&&dep[v]==-1){
                dep[v]=dep[u]+1;
                if(v==n+1)return 1;
                else q.push(v);
            }
        }
    }
    return 0;
}
int Dinic(){
    int maxf=0;vector<int>q;
    while(BFS()){
        memset(vis,0,sizeof(vis));
        q.push_back(0);vis[0]=1;
        while(!q.empty()){
            int p=q.back();
            if(p==n+1){
                int minn,minx=0x7fffffff;
                for(int i=1;i<q.size();i++){
                    int u=q[i-1],v=q[i];
                    if(map[u][v]<minx){
                        minx=map[u][v];
                        minn=u;
                    }
                }
                maxf+=minx;
                for(int i=1;i<q.size();i++){
                    int u=q[i-1],v=q[i];
                    map[u][v]-=minx;map[v][u]+=minx;
                }
                while(!q.empty()&&q.back()!=minn){
                    vis[q.back()]=0;q.pop_back();
                }
            }
            else {
                int i;
                for(i=0;i<=n+1;i++){
                    if(map[p][i]>0&&dep[i]==dep[p]+1
                        &&!vis[i]){
                        q.push_back(i);vis[i]=1;
                        break;
                    }
                }
                if(i>n+1)q.pop_back();
            }
        }
    }
    return maxf;
}
int main(){
    cin>>m>>n;
    for(int i=1;i<=m;i++)
      scanf("%d",pig+i);
    
    int num,k;
    for(int i=1;i<=n;i++){
        scanf("%d",&num);
        for(int j=0;j<num;j++){
            scanf("%d",&k);
            if(house[k]==0) map[0][i]+=pig[k];
            else map[house[k]][i]=INF;
            house[k]=i;
        }
        int tpig;
        scanf("%d",&tpig);
        map[i][n+1]=tpig;
    }
    cout<<Dinic()<<endl;
    return 0;
}