poj1149 PIGS

PIGS

Time Limit: 1000MSMemory Limit: 10000KTotal Submissions: 13677Accepted: 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

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

    这题难点在于如何建图，如何理解题目中的“if Mirko wants, he can redistribute the remaining pigs across the unlocked pig-houses.”也将成为建图的关键。

    如果我们把每个顾客作为图中的点，则对于拥有某个猪圈的钥匙的连续的两个顾客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拥有猪圈1,2，3的钥匙，j拥有猪圈1的钥匙，这样i,j同时拥有3号猪圈的钥匙，按理说j只能打开1号猪圈，但i顾客购买猪后，Mirko是可以随意分配1,2,3中猪的数量的，所以j顾客实质上是可以得到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，分别与n个顾客建边，鉴于上面的建边方式，s应与每个猪圈的第一个顾客建边，边容量为猪圈中猪的数量，若某个顾客同时为多个猪圈的第一个顾客，则数量相加。

    每个顾客购买猪的数量是有上限的，所以我们应该引入一个汇点t，每个顾客与t建一条边，边容量为每个顾客需要购买的猪的数量。

#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];//每个猪圈的前一个顾客    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;}