poj1149 PIGS(网络流)

PIGS

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 16859 Accepted: 7563

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 33 1 102 1 2 22 1 3 31 2 6

Sample Output

7

Source

Croatia OI 2002 Final Exam - First day

/*郭炜的暑期资料讲的挺详细的。。构图难，难于上青天。....万能的网络流Time:2014-12-19 12:37*/#include<cstdio>#include<cstring>#include<queue>#include<algorithm>using namespace std;const int INF=1<<28;const int MAX=1010;int n,m;int g[MAX][MAX];int pigs[MAX],d[MAX],house[MAX];bool BFS(int s,int t){    queue<int>q;    memset(d,-1,sizeof(d));    q.push(s);d[s]=0;    while(!q.empty()){        int u=q.front();q.pop();        for(int v=1;v<=n+1;v++){            if(g[u][v]>0&&d[v]==-1){                d[v]=d[u]+1;                q.push(v);            }        }    }    return d[t]!=-1;}int DFS(int u,int cur_flow){    if(u==n+1) return cur_flow;    int temp=cur_flow;    for(int v=0;v<=n+1;v++){        if(g[u][v]>0&&d[u]+1==d[v]){            int flow=DFS(v,min(temp,g[u][v]));            g[u][v]-=flow;            g[v][u]+=flow;            temp-=flow;        }    }    return cur_flow-temp;}int Dinic(int s,int t){    int flow=0;    while(BFS(s,t)){            int temp;        while(temp=DFS(s,INF)){            flow+=temp;        }    }    return flow;}int main(){    while(scanf("%d%d",&m,&n)!=EOF){        int sour=0,des=n+1;        for(int i=1;i<=m;i++){            scanf("%d",&pigs[i]);//i个猪圈里边猪的个数        }        memset(g,0,sizeof(g));        memset(house,0,sizeof(house));        int keys,id;        for(int i=1;i<=n;i++){            scanf("%d",&keys);            while(keys--){                scanf("%d",&id);//猪圈编号                if(!house[id]){//第一次打开                    g[sour][i]+=pigs[id];                }else{                    g[house[id]][i]=INF;                }                house[id]=i;//维护上一个打开猪圈的顾客编号            }            scanf("%d",&g[i][des]);//顾客到终点连一条边        }        printf("%d\n",Dinic(sour,des));    }return 0;}