POJ 1149 PIGS

PIGS

Time Limit: 1000MS

Memory Limit: 10000KB

64bit IO Format: %I64d & %I64u

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

建图思路很好

此处搬运自AaronPolaris的博客:

　　构图方式：

　　①把每个顾客看作除源点和汇点以外的节点。

　　②从源点向每个猪圈的第一个顾客连一条边，容量为该猪圈最初的猪的数量。

　　③每个猪圈的前后两个顾客之间连一条边，容量为正无穷。因为可以任意分配每个猪圈中的猪的数量。

　　④从每个顾客向汇点连一条边，容量为要购买的猪的数量。

搬运完毕

//POJ - 1149 PIGS
/*by SilverN*/
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
using namespace std;
const int INF=0xfffff;
const int mxn=420;//最大顾客数 
const int mxm=1200;//最大猪圈数
int s,t;
int w[mxn][mxn];//容量
int d[mxn];//深度 
int house[mxm];//猪圈里猪数量 
int last[mxn];
int M,N;//猪圈数，顾客数 

void init(){
    int i,j;

    scanf("%d%d",&M,&N);
    int num;
    int x;
    s=0;t=N+1;
    for(i=1;i<=M;i++)scanf("%d",&house[i]);
    for(i=1;i<=N;i++){
        scanf("%d",&num);
        for(j=1;j<=num;j++){
            scanf("%d",&x);
            if(!last[x])w[s][i]+=house[x];
            //如果是来到此猪圈的第一个顾客,从源点连边到顾客点，容量为猪圈里猪的数量 
            else w[last[x]][i]=INF;
            //在同一个猪圈的前后顾客之间连边 
            last[x]=i;
        }
        scanf("%d",&x);
        w[i][t]=x;//顾客到汇点连边，容量为欲购买量 
    }
    return;
}
bool BFS(int s){
    queue<int>q;
    memset(d,-1,sizeof(d));
    q.push(s);
    d[s]=0;
    int i;
    while(!q.empty()){
        int u=q.front();
        if(u==t)return true;
        q.pop();
        for(i=0;i<=t;i++){
            if(w[u][i] && d[i]==-1){
                d[i]=d[u]+1;
                q.push(i);
            }
        }
    }
    return false;
}
int DFS(int x,int low){
    if(x==t)return low;
    int i,a;
    for(i=0;i<=t;i++){
        if(w[x][i]>0 && d[i]==d[x]+1)
          if(a=DFS(i,min(w[x][i],low))){
              w[x][i]-=a;
              w[i][x]+=a;
              return a;
          }
    }
    return 0;
}
void dinic(){
    int ans=0;
    int flow=0;
    while(BFS(s)){
        while(flow=DFS(s,INF)){
            ans+=flow;
        }
    }
    printf("%d
",ans);
    return;
}
int main(){
    init();
    dinic();
    return 0;
}