数量queuepoj1149 PIGS

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PIGS

    

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13677		Accepted: 6044

    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

    每日一道理
这浓浓的母爱使我深深地认识到：即使你是一只矫健的雄鹰，也永远飞不出母爱的长空；即使你是一条扬帆行驶的快船，也永远驶不出母爱的长河！在人生的路上不管我们已走过多远，还要走多远，我们都要经过母亲精心营造的那座桥！

    

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

这题难点在于如何建图，如何懂得题目中的“if Mirko wants, he can redistribute the remaining pigs across the unlocked pig-houses.”也将成为建图的症结。

如果我们把每个主顾作为图中的点，则对于具有某个猪圈的钥匙的连续的两个主顾i和j,建一条边i->j。现在解释为何可以这样建边：比如说i具有猪圈1,2，3的钥匙，j具有猪圈1的钥匙，这样i,j同时具有3号猪圈的钥匙，按理说j只能打开1号猪圈，但i主顾购买猪后，Mirko是可以随便分配1,2,3中猪的数量的，所以j主顾实质上是可以得到1,2,3中的猪的。

因为猪圈有m个，所以我们可以加一个源点s，分离与n个主顾建边，鉴于下面的建边方式，s应与每个猪圈的第一个主顾建边，边容量为猪圈中猪的数量，若某个主顾同时为多个猪圈的第一个主顾，则数量相加。

每个主顾购买猪的数量是有上限的，所以我们应该引入一个汇点t，每个主顾与t建一条边，边容量为每个主顾须要购买的猪的数量。

#include <iostream>
#include<cstdio>

using namespace std;
const int MAXN=105;
const int INF=(1<<29);
int flow[MAXN][MAXN];//容量限制
int dalta[MAXN];//改变量
int pre[MAXN];
int flag[MAXN];//标号
int m,n;

int EK()
{
    int i,maxflow=0;
    int queue[MAXN],front,rear;
    while(1)
    {
        front=rear=0;
        for(i=1;i<=n+1;i++)
            flag[i]=0;
        flag[0]=1;
        pre[0]=0;
        dalta[0]=INF;
        queue[rear++]=0;
        while(front!=rear&&!flag[n+1])
        {
            int v=queue[front++];
            for(i=1;i<=n+1;i++)
            {
                if(flag[i])
                    continue;
                if(flow[v][i])
                {
                    dalta[i]=min(dalta[v],flow[v][i]);
                    flag[i]=1;
                    pre[i]=v;
                    queue[rear++]=i;
                }
            }
        }
        if(!flag[n+1])
            break;
        maxflow+=dalta[n+1];
        i=n+1;
        while(i!=0)
        {
            flow[pre[i]][i]-=dalta[n+1];
            flow[i][pre[i]]+=dalta[n+1];
            i=pre[i];
        }
    }
    return maxflow;
}

int main()
{
    int i,j;
    int pigs[1005];//每个猪圈中的猪的数量
    int before[1005];//每个猪圈的前一个主顾
    while(~scanf("%d%d",&m,&n))
    {
        for(i=0;i<=n+1;i++)
            for(j=0;j<=n+1;j++)
                flow[i][j]=0;
        for(i=1;i<=m;i++)
        {
            scanf("%d",pigs+i);
            before[i]=-1;
        }
        for(i=1;i<=n;i++)
        {
            int num;
            scanf("%d",&num);
            while(num--)
            {
                scanf("%d",&j);
                if(before[j]==-1)
                {
                    before[j]=i;
                    flow[0][i]+=pigs[j];
                }
                else
                {
                    flow[before[j]][i]=INF;
                    before[j]=i;
                }
            }
            scanf("%d",&j);
            flow[i][n+1]=j;
        }
        printf("%d\n",EK());
    }
    return 0;
}

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