网络流最大流--PIGS

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PIGS

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 12372		Accepted: 5476

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

Sample Output

POJ:http://poj.org/problem?id=1149

题意：M个猪圈，N个顾客，每个顾客有一些的猪圈的钥匙，只能购买这些有钥匙的猪圈里的猪，而且要买一定数量的猪，每个猪圈有已知数量的猪，
但是猪圈可以重新打开，将猪的个数，重新分配，以达到卖出的猪的数量最多。

思路：刚学网络流，表示很菜很菜很菜~~
①构造网络，将顾客看成源点和汇点以外的结点，并设另外两个节点：源点和汇点。
②源点和每个猪圈的第一个顾客连边，边的权是开始时候猪圈中猪的数量。
③ 若源点和某个节点之间有重边，则将权合并
④顾客j紧跟顾客i之后打开某个猪圈，则<i.j>的权是正无穷。
⑤每个顾客和会点之间连边，边的权值是顾客所希望购买的猪的数量。
例如：样例中的就可以建立如图：

其中inf是正无穷~~

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 300000000
#define M 1002
#define N 102
int s,t;
int cus[N][N];
int flow[N][N];
int i,j;
void ford()
{
    int prev[N];
    int minflow[N];
    int queue[N];
    int qs,qe,v,p;
    for(i=0;i<N;i++)
       for(j=0;j<N;j++)
       flow[i][j]=0;
    minflow[0]=INF;
    while(1)
    {
        for(i=0;i<N;i++)
        prev[i]=-2;
        prev[0]=-1;
        qs=0;
        queue[qs]=0;
        qe=1;
        while(qs<qe&&prev[t]==-2)
        {
            v=queue[qs++];
            for(i=0;i<t+1;i++)
            {
                if(prev[i]==-2&&(p=cus[v][i]-flow[v][i]))
                {
                    prev[i]=v; queue[qe++]=i;
                    minflow[i]=(minflow[v]<p)?minflow[v]:p;
                }
            }
        }
        if(prev[t]==-2) break;
        for(i=prev[t],j=t;i!=-1;j=i,i=prev[i])
        {
            flow[i][j]=flow[i][j]+minflow[t];
            flow[j][i]=-flow[i][j];
        }
    }
    for(i=0,p=0;i<t;i++)
      p=p+flow[i][t];
    printf("%d\n",p);
}
int main()
{
    int m,n;
    int num,k;
    int house[M],last[M];
    while(scanf("%d %d",&m,&n)!=EOF)
    {
        memset(last,0,sizeof(last));
        memset(cus,0,sizeof(cus));
        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=0;j<num;j++)
            {
                scanf("%d",&k);
                if(last[k]==0)
                   cus[s][i]=cus[s][i]+house[k];
                else
                   cus[last[k]][i]=INF;
                last[k]=i;
            }
            scanf("%d",&cus[i][t]);
        }
        ford();
    }
}

poj1273 一般增光路算法（残余网络），find_augment_path求增光路，并通过augment_flow计算可改进量，然后调用update_flow更新网络流，直到不存在增光路为止。

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
#define M 210
struct Matix{
    int c,f;
}edge[M][M];
int n,m;
int s,t;
int resi[M][M];
int qu[M*M],qs,qe;
int pre[M];
int vis[M];
int maxflow,min_augment;
void find_augment_path(){
    int i,v;
    memset(vis,0,sizeof(vis));
    qs=0;qu[qs]=s;
    pre[s]=s;vis[s]=1;qe=1;
    memset(resi,0,sizeof(resi));
    memset(pre,0,sizeof(pre));
    while(qs<qe&&pre[t]==0){
        v=qu[qs++];
        for(i=1;i<=n;i++){
            if(vis[i]==0){
                if(edge[v][i].c-edge[v][i].f>0){
                    resi[v][i]=edge[v][i].c-edge[v][i].f;
                    pre[i]=v;qu[qe++]=i;vis[i]=1;
                }else if(edge[i][v].f>0){
                    resi[v][i]=edge[i][v].f;
                    pre[i]=v;qu[qe++]=i;vis[i]=1;
                }
            }
        }
    }
}
void augment_flow(){
    int i=t,j;
    if(pre[i]==0) {min_augment=0;return;}
    j=0x7fffffff;
    while(i!=s){
        if(resi[pre[i]][i]<j) j=resi[pre[i]][i];
        i=pre[i];
    }
    min_augment=j;
}
void update_flow(){
    int i=t;
    if(pre[i]==0) return ;
    while(i!=s){
        if(edge[pre[i]][i].c-edge[pre[i]][i].f>0)
            edge[pre[i]][i].f+=min_augment;
        else if(edge[i][pre[i]].f>0) edge[pre[i]][i].f+=min_augment;
        i=pre[i];
    }
}
void solve(){
    s=1;t=n;
    maxflow=0;
    while(1){
        find_augment_path();
        augment_flow();
        maxflow+=min_augment;
        if(min_augment>0) update_flow();
        else return ;
    }
}
int main(){
    int i;
    int u,v,w;
    while(scanf("%d%d",&m,&n)!=EOF){
        memset(edge,0,sizeof(edge));
        for(i=0;i<m;i++){
            scanf("%d%d%d",&u,&v,&w);
            edge[u][v].c+=w;
        }
        solve();
        printf("%d\n",maxflow);
    }
    return 0;
}