网络流(最大流)：POJ 1149 PIGS

PIGS

Time Limit: 1000ms

Memory Limit: 10000KB

This problem will be judged on PKU.
64-bit integer(整数) IO format: %lld      Java class name: Main

 

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(关于) 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(取…最大值) the number of pigs sold.
More precisely(精确地), the procedure(程序) is as following: the customer arives, 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(重新分配) 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(投入) contains two integers(整数) 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(不减少的) ) 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(输出) 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
　　
　　建模题，这里需要注意对空间的优化。
　　题意：迈克有个养猪场，养猪场里有M个猪圈，每个猪圈都上了锁。迈克没有钥匙，而要买猪的顾客一个接一个来到养猪场，每个顾客有一些猪圈的钥匙，要买一定数量的猪。当每个顾客来时，将有钥匙的猪圈全部打开，从中挑出一些买走，然后迈克可以重新分配这些猪圈里面的猪。当顾客离开后，门又被锁上。问迈克最多可以卖多少猪。
　　建模：先从源点给每个猪圈连一条边，容量是猪圈中猪的头数。这时再添加顾客，对于每一个顾客，查找他要开的每一个猪圈，如果他要开猪圈A，那么现在分情况讨论：
　　<1>若以前(先后顺序，时间上的)没有顾客开过A猪圈，那么就连一条A到这个顾客的边，容量为INF，同时标记这个人为这个猪圈的“开启者”
　　<2>若有，则将A的“开启者”连到这个人，容量为INF
　　最后每个顾客连边到汇点，容量为各自的需求，接着跑一遍最大流就可以啦，这里我用了ISAP算法

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>

using namespace std;
const int INF=2147483647;
const int maxn=1010,maxm=4010;
int cnt,fir[maxn],nxt[maxm],cap[maxm],to[maxm],dis[maxn],gap[maxn],path[maxn],used[maxn];

void addedge(int a,int b,int c)
{
    nxt[++cnt]=fir[a];
    to[cnt]=b;
    cap[cnt]=c;
    fir[a]=cnt;
}

bool BFS(int S,int T)
{
    memset(dis,0,sizeof(dis));
    dis[T]=1;
    queue<int>q;q.push(T);
    while(!q.empty())
    {
        int node=q.front();q.pop();
        for(int i=fir[node];i;i=nxt[i])
        {
            if(dis[to[i]])continue;
            dis[to[i]]=dis[node]+1;
            q.push(to[i]);
        }
    }
    return dis[S];
}
int fron[maxn];
int ISAP(int S,int T)
{
    if(!BFS(S,T))
        return 0;
    for(int i=1;i<=T;i++)++gap[dis[i]];
    int p=S,ret=0;
    memcpy(fron,fir,sizeof(fir));
    while(dis[S]<=T)
    {
        if(p==T){
            int f=INF;
            while(p!=S){
                f=min(f,cap[path[p]]);
                p=to[path[p]^1];
            }
            p=T;ret+=f;
            while(p!=S){
                cap[path[p]]-=f;
                cap[path[p]^1]+=f;
                p=to[path[p]^1];
            }
        }
        int &ii=fron[p];
        for(;ii;ii=nxt[ii]){
            if(!cap[ii]||dis[to[ii]]+1!=dis[p])
                continue;
            else 
                break;
        }        
        if(ii){
            p=to[ii];
            path[p]=ii;
        }
        else{
            if(--gap[dis[p]]==0)break;
            int minn=T+1;
            for(int i=fir[p];i;i=nxt[i])
                if(cap[i])
                    minn=min(minn,dis[to[i]]);
            gap[dis[p]=minn+1]++;
            fron[p]=fir[p];
            if(p!=S)
                p=to[path[p]^1];        
        }
    }
    return ret;
}

void Init()
{
    memset(fir,0,sizeof(fir));
    memset(used,0,sizeof(used)); 
    cnt=1;
}
int main()
{
    int n,m,num,k,need;
    while(~scanf("%d%d",&m,&n))
    {
        Init();
        for(int i=1;i<=m;i++){
            scanf("%d",&num);
            addedge(0,i,num);
            addedge(i,0,0);
        }
        for(int i=m+1;i<=m+n;i++){
            scanf("%d",&k);
            while(k--){
                scanf("%d",&num);
                if(used[num]){
                    addedge(used[num],i,INF);
                    addedge(i,used[num],0);
                }
                else{
                    used[num]=i;
                    addedge(num,i,INF);
                    addedge(i,num,0);
                }
                
            }
            scanf("%d",&need);
            addedge(i,n+m+1,need);
            addedge(n+m+1,i,0);
        }
        printf("%d
",ISAP(0,n+m+1));
    }
    return 0;
}