PIGS

PIGS
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 18209 Accepted: 8277

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
题意:有不同的猪圈,买家手里有着猪圈的钥匙,拥有那个猪圈的钥匙就可以买这个猪圈的猪,怎样安排买猪才能卖出更多的猪;

#include <map>#include <list>#include <climits>#include <cmath>#include <queue>#include <stack>#include <string>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;#define eps 1e-9#define LL long long#define PI acos(-1.0)#define INF 0x3f3f3f3f#define CRR fclose(stdin)#define CWW fclose(stdout)#define RR freopen("input.txt","r",stdin)#define WW freopen("output.txt","w",stdout)const int Max = 100000;int cus[110][110];//图的容量int flow[110][110];//图的流量int num[1100];//猪的数量int pre[1100];//前一个买家int n,m,data;int t,s;int ford()//一般增广路算法{    int minflow[110];    queue<int >Q;    memset(flow,0,sizeof(flow));    minflow[0] = INF;    t=n+1;    s=0;    while(1)    {        for(int i=0;i<=t;i++)        {            pre[i]=-2;        }        while(!Q.empty())        {            Q.pop();        }        pre[0]=-1;        Q.push(0);        while(!Q.empty()&&pre[t]==-2)//BFS找增广路        {            int v=Q.front();            int p;            Q.pop();            for(int i=0;i<=t;i++)            {                if(pre[i]==-2&&(p=cus[v][i]-flow[v][i]))                {                    pre[i]=v;                    Q.push(i);                    minflow[i]=min(minflow[v],p);                }            }        }        if(pre[t]==-2)//如果找不到汇点,则说明图中不存在增广路,算法结束        {            break;        }        for(int i=pre[t],j=t;i!=-1;j=i,i=pre[i])//回溯进行增广        {            flow[i][j] +=minflow[t];            flow[j][i]-=minflow[t];        }    }    int sum=0;    for(int i=0;i<t;i++)    {        sum+=flow[i][t];//每个点到汇点的总和为最大流    }    return sum;}int main(){    while(~scanf("%d %d",&m,&n))    {        memset(pre,-1,sizeof(pre));        memset(cus,0,sizeof(cus));        for(int i=1;i<=m;i++)        {            scanf("%d",&num[i]);        }        int s;        for(int i=1;i<=n;i++)        {            scanf("%d",&data);            for(int j=0;j<data;j++)            {                scanf("%d",&s);                if(pre[s]==-1)//如果这个猪圈之前没有人,则这个人是第一个人,所以与源点之间的容量为猪圈的猪的数量                {                    cus[0][i]+=num[s];                }                else                {                    cus[pre[s]][i]=INF;//如果已经有人买了,所以他必须在这个人之后买,但两人之间又没有关系,所以容量为INF(无穷大)                }                pre[s]=i;            }            scanf("%d",&cus[i][n+1]);//到汇点的容量为每个人想买猪的数量        }        printf("%d\n",ford());    }    return 0;}