poj--1149--PIGS（最大流经典建图）

PIGS

Time Limit: 1000MS

Memory Limit: 10000KB

64bit IO Format: %I64d & %I64u

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Status

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



最大流经典建图，有m个猪圈，n个顾客，m个猪圈的钥匙在n个顾客手中，n个顾客一次来卖猪，当他们打开过猪圈之后农场主可以将这些猪圈里的猪按照自己的意愿进行转移，求这n个顾客可以买到多少头猪



建立超级源点与超级汇点，超级源点指向每个猪圈到达的第一个顾客，随后的顾客与前一个顾客产生一条正无穷的边权，而从源点出来的边权就是该猪圈的猪的数目，因为猪圈打开过之后猪就可以转移，那么随后的边权就应该是无穷，因为之后这个猪圈中猪的数目是不确定的，但是超级汇点要与每一个顾客建边，边权就是该顾客的需求量



可以参照大牛的分析，非常详细

http://wenku.baidu.com/view/0ad00abec77da26925c5b01c.html

#include<stdio.h>
#include<string.h>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
#define MAXN 2000
#define MAXM 50000
#define INF 10000000
struct node
{
	int u,v,cap,flow,next;
}edge[MAXM];
int cur[MAXN],dis[MAXN],vis[MAXN];
int head[MAXN],pig[MAXN],want[MAXN];
vector<int>first[MAXN];
int m,n,top;
void init()
{
	top=0;
	memset(head,-1,sizeof(head));
	for(int i=0;i<=m;i++)
	first[i].clear();
}
void add(int u,int v,int w)
{
	int i=-1;
	for(i=head[u];i!=-1;i=edge[i].next)//判断是否有重边
	{
		if(edge[i].v==v)
		break;
	}
	if(i==-1)//没有重边的话直接建图
	{
		node E1={u,v,w,0,head[u]};
		edge[top]=E1;
		head[u]=top++;
		node E2={v,u,0,0,head[v]};
		edge[top]=E2;
		head[v]=top++;
	}
	else//重边就流量汇总
	{
		if(w!=INF) edge[i].cap+=w;
	}
}
void getmap()
{
	for(int i=1;i<=m;i++)
	{
		int u=first[i][0];
		add(0,u,pig[i]);//超级源点与每个猪圈进行联系，边权为猪圈猪的数目
		for(int j=0;j<first[i].size()-1;j++)
		{
			int x=first[i][j];//将第一个顾客与后来的顾客产生联系
			int y=first[i][j+1];
			add(x,y,INF);
		}
	}
	for(int i=1;i<=n;i++)
	add(i,n+1,want[i]);//超级汇点与每一个顾客链接
}
void input()
{
	int t,a;
	for(int i=1;i<=m;i++)
	scanf("%d",&pig[i]);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&t);
		while(t--)
		{
			scanf("%d",&a);
			first[a].push_back(i);//first[a][0]就是第一个顾客
		}
		scanf("%d",&want[i]);
	}
}
bool bfs(int s,int e)
{
	memset(vis,0,sizeof(vis));
	memset(dis,-1,sizeof(dis));
	queue<int>q;
	while(!q.empty()) q.pop();
	q.push(s);
	vis[s]=1;
	dis[s]=0;
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		for(int i=head[u];i!=-1;i=edge[i].next)
		{
			node E=edge[i];
			if(!vis[E.v]&&E.cap>E.flow)
			{
				vis[E.v]=1;
				dis[E.v]=dis[E.u]+1;
				if(E.v==e) return true;
				q.push(E.v);
			}
		}
	}
	return false;
}
int dfs(int x,int a,int e)
{
	if(a==0||x==e)
	return a;
	int flow=0,f;
	for(int i=cur[x];i!=-1;i=edge[i].next)
	{
		node& E=edge[i];
		if(dis[x]+1==dis[E.v]&&(f=dfs(E.v,min(a,E.cap-E.flow),e))>0)
		{
			E.flow+=f;
			edge[i^1].flow-=f;
			flow+=f;
			a-=f;
			if(a==0) break;
		}
	}
	return flow;
}
int MAXflow(int s,int e)
{
	int flow=0;
	while(bfs(s,e))//是否可以增广
	{
		memcpy(cur,head,sizeof(head));
		flow+=dfs(s,INF,e);	//printf("%d\n",flow);
	}
	return flow;
}
int main()
{
	while(scanf("%d%d",&m,&n)!=EOF)
	{
		init();
		input();
		getmap();
		printf("%d\n",MAXflow(0,n+1));
	}
	return 0;
}