POJ 1149 PIGS

PIGS

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 20582		Accepted: 9389

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m个猪圈，n个人，每个人可以打开一些猪圈，从这些猪圈里领猪，当前开着的猪圈中的猪可以随意走动到开着的猪圈中，每个客人走之后猪圈会再次关上，每个客人有期望数量，问最多可以领走多少小猪^(*￣(oo)￣)^

分析：

“这水题写什么写”---来自YSQ的嘲讽TAT...

确实没什么好写的...

不过有一点思想很重要就是可以用+∞的边表示流量传递...

对于每个猪圈我们从S向每个猪圈连一条容量为猪数量的边，从每个顾客到T连一条容量为期望领养数量的边，对于每一个客人，如果这个客人是第一次打开这个猪圈的客人，那么猪圈向这个客人连+∞的边，否则上一个打开这个猪圈的客人向当前客人连一条+∞的边...

代码：

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
//by NeighThorn
#define inf 0x3f3f3f3f
using namespace std;
//大鹏一日同风起，扶摇直上九万里

const int maxn=2000+5,maxm=800000+5;

int n,m,S,T,cnt,hd[maxn],to[maxm],fl[maxm],nxt[maxm],pre[maxn],pos[maxn],num[maxn];

inline void add(int s,int x,int y){
    fl[cnt]=s;to[cnt]=y;nxt[cnt]=hd[x];hd[x]=cnt++;
    fl[cnt]=0;to[cnt]=x;nxt[cnt]=hd[y];hd[y]=cnt++;
}

inline bool bfs(void){
    memset(pos,-1,sizeof(pos));
    int head=0,tail=0,q[maxn];
    q[0]=S;pos[S]=0;
    while(head<=tail){
        int top=q[head++];
        for(int i=hd[top];i!=-1;i=nxt[i])
            if(pos[to[i]]==-1&&fl[i])
                pos[to[i]]=pos[top]+1,q[++tail]=to[i];
    }
    return pos[T]!=-1;
}

inline int find(int v,int f){
    if(v==T)
        return f;
    int res=0,t;
    for(int i=hd[v];i!=-1&&f>res;i=nxt[i])
        if(pos[to[i]]==pos[v]+1&&fl[i])
            t=find(to[i],min(fl[i],f-res)),res+=t,fl[i]-=t,fl[i^1]+=t;
    if(!res)
        pos[v]=-1;
    return res;
}

inline int dinic(void){
    int res=0,t;
    while(bfs())
        while(t=find(S,inf))
            res+=t;
    return res;
}

signed main(void){
    scanf("%d%d",&m,&n);cnt=0;
    S=0,T=n+m+1;memset(hd,-1,sizeof(hd));
    for(int i=1;i<=m;i++)
        scanf("%d",&num[i]),pre[i]=i,add(num[i],S,i);
    for(int i=1;i<=n;i++){
        int x;scanf("%d",&x);
        for(int j=1,y;j<=x;j++)
            scanf("%d",&y),add(inf,pre[y],i+m),pre[y]=i+m;
        scanf("%d",&x);add(x,i+m,T);
    }
    printf("%d
",dinic());
    return 0;
}//Cap ou pas cap. Cap;

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By NeighThorn