poj 1149 PIGS

PIGS

http://poj.org/problem?id=1149

Time Limit: 1000MS		Memory Limit: 10000K

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

题目大意：

Mirko 养着一些猪, 猪关在一些猪圈里面, 猪圈是锁着的他自己没有钥匙 (汗)
只有要来买猪的顾客才有钥匙, 顾客依次来每个顾客会用他的钥匙打开一些猪圈买
走一些猪, 然后锁上.
在锁上之前 Mirko 有机会重新分配这几个已打开猪圈的猪, 现在给出一开始每个猪
圈的猪数, 每个顾客所有的钥匙和要买走的猪数问 Mirko 最多能卖掉几头猪

以样例输入为例

有m个猪圈，n个顾客

接下来一行有m个数，表示每个猪圈开始猪的数量

接下来n行，

每行第一个数表示第i个顾客手里的钥匙数p，后面p个数表示是哪个猪圈的钥匙，最后一个数表示他要买的猪的数量

最大流

很容易想到的建图方法

 

上图修改如下：

 

但是这样会有100000左右个点，TLE

所以我们要对点进行合并

合并规则：

规律 1. 如果几个节点的流量的来源完全相同，则可以把它们合并成一个。

 规律 2. 如果几个节点的流量的去向完全相同，则可以把它们合并成一个。
  规律 3. 如果从点 u 到点 v 有一条容量为 +∞ 的边，并且 u 是 v 的唯一流量来源，或者 v 是 u 的唯一流量去向，则可以把 u 和 v 合并成一个节点。

所以最终构图方式：

1、由源点向每个猪圈第一个要访问的顾客连一条边，流量为一开始猪的数量，如果源点向同一个顾客有多条边，就把他们合并为1条边，流量相加。

2、如果顾客j在顾客i后访问猪圈k，则由i向j连一条流量为inf的边

3、每个顾客向汇点连一条边，流量为他要买的猪的数量

更详细的解释推荐博客http://blog.csdn.net/shahdza/article/details/6831946

#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#include<iostream>
#define inf 600001
using namespace std;
int n,m,qq,tot=1,ans;
int src,decc;
int front[112],cap[22001],lev[112],cnt[112],nextt[22001],to[22001];
queue<int>q;
int pre[1002],buy[112],kai[112][112],last[1002];
void add(int u,int v,int w)
{
    to[++tot]=v;cap[tot]=w;nextt[tot]=front[u];front[u]=tot;
    to[++tot]=u;cap[tot]=0;nextt[tot]=front[v];front[v]=tot;
}
bool bfs()
{
    for(int i=0;i<=m+1;i++) {cnt[i]=front[i];lev[i]=-1;}
    while(!q.empty()) q.pop();
    q.push(src);lev[src]=0;
    while(!q.empty())
    {
        int now=q.front();q.pop();
        for(int i=front[now];i!=0;i=nextt[i])
        {
            int t=to[i];
            if(cap[i]>0&&lev[t]==-1)
            {
                q.push(t);
                lev[t]=lev[now]+1;
                if(t==decc) return true;
            }
        }
    }
    return false;
}
int dinic(int now,int flow)
{
    if(now==decc) return flow;
    int delta,rest=0;
    for(int & i=cnt[now];i!=0;i=nextt[i])
    {
        int t=to[i];
        if(lev[t]==lev[now]+1&&cap[i]>0)
        {
            delta=dinic(t,min(cap[i],flow-rest));
            if(delta)
            {
                cap[i]-=delta;cap[i^1]+=delta;
                rest+=delta;if(rest==flow) break;
            }
        }
    }
    if(rest!=flow) lev[now]=-1;
    return rest;
}
int main()
{
   scanf("%d%d",&n,&m);  
   int p,x;
   for(int i=1;i<=n;i++) scanf("%d",&pre[i]);
   for(int i=1;i<=m;i++)
   {
           scanf("%d",&p);
           for(int j=1;j<=p;j++) 
           {
               scanf("%d",&x);
               if(!last[x])
               {
                   kai[0][i]+=pre[x];
                   last[x]=i;
            }
            else kai[last[x]][i]=inf;
        }
        scanf("%d",&buy[i]);
   }
    decc=m+1;
    for(int i=0;i<=m;i++)
     for(int j=0;j<=m;j++)
      if(kai[i][j])     add(i,j,kai[i][j]);
    for(int i=1;i<=m;i++) add(i,decc,buy[i]);
    while(bfs()) 
     ans+=dinic(src,inf);
    printf("%d",ans);

}