POJ 1149 PIGS

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

 

题解：赤裸裸的网络流分配问题。直接建模好了，于是我们就得到了这张图：

 

然后我们发现一共有100000个节点，又不是平面图，咱们就是上最快的外国大佬的O(NM)的网络流套LCT也跑不过啊= =

那么我们要缩点了，缩点时遵循以下三条原则：

那么我们来缩点，首先最后没有开的房子（也就是红色的节点）不用建，其次蓝色的四个节点可以缩成一个点。

然后后面的好多房子也可以各种缩点。。。然后图就变成这个德行了：

 

新的建图规则可以总结一下：

 

然后水过就好= =

 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<queue>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=100+10,maxm=1000+10,maxe=200000+10;
int n,m,S,T;
int cus[maxm][maxn],pig[maxm],buy[maxn],num[maxm];
struct Tedge{int x,y,flow,cap,next;};
struct Dinic{
    Tedge adj[maxe];int fch[maxn],ms;
    int n,m,S,T,d[maxn],cur[maxn];bool vis[maxn];
    void init(int n){
        this->n=n;
        ms=0;
        memset(fch,-1,sizeof(fch));
        return;
    }
    void AddEdge(int a,int b,int c){
        adj[ms]=(Tedge){a,b,0,c,fch[a]};fch[a]=ms++;
        adj[ms]=(Tedge){b,a,0,0,fch[b]};fch[b]=ms++;
        return;
    }
    bool BFS(){
        memset(vis,false,sizeof(vis));
        queue<int>Q;Q.push(S);vis[S]=true;d[S]=0;
        while(!Q.empty()){
            int u=Q.front();Q.pop();
            for(int i=fch[u];i!=-1;i=adj[i].next){
                int v=adj[i].y;
                if(!vis[v]&&adj[i].cap>adj[i].flow){
                    vis[v]=true;
                    d[v]=1+d[u];
                    Q.push(v);
                }
            }
        } return vis[T];
    }
    int DFS(int u,int a){
        if(u==T||!a) return a;
        int flow=0,f;
        for(int& i=cur[u];i!=-1;i=adj[i].next){
            int v=adj[i].y;
            if(d[v]==d[u]+1&&(f=DFS(v,min(a,adj[i].cap-adj[i].flow)))>0){
                flow+=f;
                a-=f;
                adj[i].flow+=f;
                adj[i^1].flow-=f;
                if(!a) break;
            }
        } return flow;
    }
    int MaxFlow(int S,int T){
        this->T=T;this->S=S;
        int flow=0;
        while(BFS()){
            for(int i=0;i<n;i++) cur[i]=fch[i];
            flow+=DFS(S,inf);
        } return flow;
    }
}sol;
inline int read(){
    int x=0,sig=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-') sig=-1;ch=getchar();}
    while(isdigit(ch)) x=10*x+ch-'0',ch=getchar();
    return x*=sig;
}
inline void write(int x){
    if(x==0){putchar('0');return;} if(x<0) putchar('-'),x=-x;
    int len=0,buf[15]; while(x) buf[len++]=x%10,x/=10;
    for(int i=len-1;i>=0;i--) putchar(buf[i]+'0');return;
}
void init(){
    m=read();n=read();
    for(int i=0;i<m;i++) pig[i]=read();
    for(int i=0;i<n;i++){
        int a=read(),b;
        for(int j=0;j<a;j++){
            b=read();
            b--;
            cus[b][num[b]++]=i;
        }
        buy[i]=read();
    }
    return;
}
void work(){
    
    return;
}
void print(){
    S=n;T=n+1;
    sol.init(n+2);
    for(int i=0;i<n;i++) sol.AddEdge(i,T,buy[i]);
    for(int i=0;i<m;i++){
        if(num[i]>0) sol.AddEdge(S,cus[i][0],pig[i]);
        for(int j=1;j<num[i];j++) sol.AddEdge(cus[i][j-1],cus[i][j],inf);
    }
    write(sol.MaxFlow(S,T));
    return;
}
int main(){
    init();work();print();return 0;
}