poj1149PIGS【最大流】

PIGS

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 20091 Accepted: 9199

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

引用黄学长的题解——

把每个卖猪的人作为点，当前来买猪的人如果先前买猪的人打开相同的猪圈，则先前那个人可以把所有能开的猪圈里的猪留给它，也就是在他们之间连一条无穷大的边，然后，如果是第一个开某个猪圈的人，就在源点与某人直接连上容量为猪圈中猪的数量的边，每个人与汇连上一条容量为它想买猪的数量的边，此时求最大流，最大流即答案

/**************poj11492016.8.203776K16MSC++2186B**************/#include <stdio.h>#include<cstring>#include <iostream>#include<vector>#include<algorithm>#include<cstring>using namespace std;const int oo=0x3f3f3f3f;const int mm=900000;const int mn=900000;int node ,scr,dest,edge;int ver[mm],flow[mm],Next[mm];int head[mn],work[mn],dis[mn],q[mn];void prepare(int _node,int _scr,int _dest){    node=_node,scr=_scr,dest=_dest;    for(int i=0; i<node; ++i)        head[i]=-1;    edge=0;}void addedge(int u,int v,int c){    ver[edge]=v,flow[edge]=c,Next[edge]=head[u],head[u]=edge++;    ver[edge]=u,flow[edge]=0,Next[edge]=head[v],head[v]=edge++;}bool Dinic_bfs(){    int i,u,v,l,r=0;    for(i=0; i<node; i++)        dis[i]=-1;    dis[q[r++]=scr]=0;    for(l=0; l<r; ++l)    {        for(i=head[u=q[l]]; i>=0; i=Next[i])        {            if(flow[i]&&dis[v=ver[i]]<0)            {                dis[q[r++]=v]=dis[u]+1;                if(v==dest)                    return 1;            }        }    }    return 0;}int Dinic_dfs(int u,int exp){    if(u==dest)        return exp;    for(int &i=work[u],v,tmp; i>=0; i=Next[i])        if(flow[i]&&dis[v=ver[i]]==dis[u]+1&&(tmp=Dinic_dfs(v,min(exp,flow[i])))>0)        {            flow[i]-=tmp;            flow[i^1]+=tmp;            return tmp;        }    return 0;}int Dinic_flow(){    int i,ret=0,delta;    while(Dinic_bfs())    {        for(i=0; i<node; i++)            work[i]=head[i];        while(delta=Dinic_dfs(scr,oo))            ret+=delta;    }    return ret;}int n,m;int last[mm],num[mm];int main(){   // freopen("cin.txt","r",stdin);    while(~scanf("%d%d",&m,&n))    {        prepare(n+2,0,n+1);        memset(last,0,sizeof(last));        for(int i=1;i<=m;i++)scanf("%d",&num[i]);        for(int i=1;i<=n;i++)        {            int a,x,b;            scanf("%d",&a);            while(a--)            {                scanf("%d",&x);                if(!last[x])addedge(0,i,num[x]);                else addedge(last[x],i,oo);                last[x]=i;            }            scanf("%d",&b);            addedge(i,n+1,b);        }        printf("%d\n",Dinic_flow());    }    return 0;}