POJ1149 PIGS

PIGS

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 20759		Accepted: 9488

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

 

暴力建图：
　　一看到题，最先想到的是要分层，以每个顾客为一层，该顾客可以打开的猪圈向该顾客连边，每层向下一层对应的猪圈以及能够相互之间调整的猪圈连边，求最大流。这样构图本身并没有错，但是数据范围是n<=100 m<=1000 图中的节点数会超过100000，时间上绝对不够

正确建图：

　　顾客之间连边.
　　新增源点s，汇点t，对于每个顾客来说，如果他是第一个打开某猪圈的，则从s向他连一条容量为该猪圈初始值的弧，如果与源点的流量已经不是0，则合并，如果不是第一个，则从上一个打开该猪圈的顾客向这个顾客连一条容量为+oo的弧，每个顾客向汇点连一条容量为该顾客希望购买的猪的数量的弧

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
using namespace std;
const int N=2250;
const int inf=0x3f3f3f3f;
struct edge{int v,cap,next;}e[N*8];int tot=1;
int n,m,k,S,T,cnt,head[N],last[N],dis[N],val[N],q[N*8];
void add(int x,int y,int z){
    e[++tot].v=y;e[tot].cap=z;e[tot].next=head[x];head[x]=tot;
    e[++tot].v=x;e[tot].cap=0;e[tot].next=head[y];head[y]=tot;
}
void mapping(){
    S=0,T=m+k+1;n=m;
    for(int i=1,w;i<=m;i++){
        scanf("%d",&w);
        add(S,i,w);
        last[i]=i;
    }
    for(int i=1,w,v,cas;i<=k;i++){
        n++;
        for(scanf("%d",&cas);cas--;){
            scanf("%d",&v);
            add(last[v],n,inf);
            last[v]=n;
        }
        scanf("%d",&w);
        add(n,T,w);
    }
}
bool bfs(){
    for(int i=S;i<=T;i++) dis[i]=-1;
    int h=0,t=1;dis[S]=0;q[t]=S;
    while(h!=t){
        int x=q[++h];
        for(int i=head[x];i;i=e[i].next){
            if(dis[e[i].v]==-1&&e[i].cap){
                dis[e[i].v]=dis[x]+1;
                if(e[i].v==T) return 1;
                q[++t]=e[i].v;
            }
        }
    }
    return 0;
}
int dfs(int x,int f){
    if(x==T) return f;
    int used=0,t;
    for(int i=head[x];i;i=e[i].next){
        if(e[i].cap&&dis[e[i].v]==dis[x]+1){
            t=dfs(e[i].v,min(e[i].cap,f));
            e[i].cap-=t;e[i^1].cap+=t;
            used+=t;f-=t;
            if(!f) return used;
        }
    }
    if(!used) dis[x]=-1;
    return used;
}
void dinic(){
    int ans=0;
    while(bfs()) ans+=dfs(S,inf);
    printf("%d
",ans);
}
void init(){
    tot=1;
    memset(head,0,sizeof head);
}
int main(){
    while(scanf("%d%d",&m,&k)==2){
        init();
        mapping();
        dinic();
    }
    return 0;
}