POJ 1149-PIGS（Ford-Fulkerson 标号法求网络最大流）

- 日理万妓 2022-09-25 06:23 236阅读 0赞

PIGS

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;10000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;20029</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;9178</td> 
  </tr> 
 </tbody> 
</table>

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

Source

[Croatia OI 2002 Final Exam - First day][]

## 题目意思： ##

有M个猪圈，N个顾客，给出每个猪圈中猪的数目。

每个顾客有A把钥匙，对应A个猪圈的编号，每个顾客会买B头猪。

Mark木有猪圈的钥匙，每个顾客来的时候把他们有钥匙的猪圈全部打开；而且Mark可以重新分配被打开的猪圈里面的猪。

顾客离开后，猪圈再次被锁上。

求Mark能卖出的猪的最大值。

## 解题思路： ##

Ford-Fulkerson 标号法求网络最大流。

除了顾客的N个顶点外，自己增加源点和汇点这两个点。

![20160815161157996][]

每个顾客购买的数目是连接到汇点上的容量；

源点与每个猪圈的第一个顾客连边，边的容量是开始时猪圈中猪的数目；

若源点与某个结点之间有重边，将权合并（如上图中Vs~V1，4就是合并了1和3）；

若顾客j紧跟i后面打开猪圈，那么<i，j>容量为正无穷（因为此时可以随意调整猪圈中的猪的数量）。

#include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <vector>
    #include <queue>
    #include <iomanip>
    #include <algorithm>
    #define maxn 1010
    #define INF 0xfffffff
    using namespace std;
    struct ArcType
    {
        int c,f;//容量、流量
    };
    
    ArcType edge[maxn][maxn];
    int n,m;//顶点数、弧数
    int s,t;
    int flag[maxn];//顶点状态：-1——未标号；0——已标号未检查；1——已标号已检查
    int prev[maxn];//标号的第一个分量：指明标号从哪个顶点而来，以便找出可改进量
    int alpha[maxn];//标号的第二个分量：可改进量α
    int que[maxn];//相当于BFS中的队列
    int v;//队列头元素
    int qs,qe;//队首队尾的位置
    int i,j;
    
    void ford()//标号法求网络最大流
    {
        int flow[maxn][maxn];//节点之间的流量Fij
        int prev[maxn];//可改进路径上前一个节点的标号，相当于标号的第一个分量
        int minflow[maxn];//每个顶点的可改进量α，相当于标号的第二个分量
        int que[maxn];
        int qs,qe;//队列首尾位置坐标
        int v,p;//当前顶点、保存Cij-Fij
        for(i=0; i<maxn; ++i)
            for(j=0; j<maxn; ++j)
                flow[i][j]=0;
        minflow[0]=INF;//源点标号的第二分量为无穷大
        while(1)//标号法
        {
            for(i=0; i<maxn; ++i)//每次标号前，每个顶点重新回到未标号状态
                prev[i]=-2;
            prev[0]=-1;
            qs=0;
            que[qs]=0;//源点入队
            qe=1;
            while(qs<qe&&prev[t]==-2)
            {
                v=que[qs];//取队列头节点
                ++qs;
                for(i=0; i<t+1; ++i)//prev[i]==-2表示顶点i未标号
                    if(prev[i]==-2&&(p=edge[v][i].c-flow[v][i]))//edge[v][i].c-flow[v][i]!=0能保证i是v邻接顶点且能进行标号
                    {
                        prev[i]=v;
                        que[qe]=i;
                        ++qe;
                        minflow[i]=(minflow[v]<p)?minflow[v]:p;
                    }
            }
            if(prev[t]==-2) break;//汇点t无标号，标号法结束
            for(i=prev[t],j=t; i!=-1; j=i,i=prev[i])//调整过程
            {
                flow[i][j]+=minflow[t];
                flow[j][i]=-flow[i][j];
            }
        }
        for(i=0,p=0; i<t; ++i)//统计进入汇点的流量即最大流的流量
            p+=flow[i][t];
        cout<<p<<endl;
    }
    int main()
    {
        ios::sync_with_stdio(false);
        cin.tie(0);
        cin>>m>>n;//顶点个数、弧数
        s=0,t=n+1;//源点和汇点
        int pig[maxn],last[maxn];//每个猪圈中猪的数量、每个猪圈前一个顾客的序号
        memset(last,0,sizeof(last));
        for(i=1; i<=m; ++i)
            cin>>pig[i];//输入每个猪圈中猪的数量
        for(i=1; i<=n; ++i)
        {
            int num;
            cin>>num;//拥有的猪圈钥匙数量
            for(j=0; j<num; ++j)
            {
                int k;
                cin>>k;//钥匙编号
                if(last[k]==0)
                    edge[s][i].c+=pig[k];
                else edge[last[k]][i].c=INF;
                last[k]=i;
            }
            cin>>edge[i][t].c;
            edge[i][t].f=0;
        }
        n+=2;//加上源点和汇点
        ford();
        return 0;
    }
    /*
    3 3
    3 1 10
    2 1 2 2
    2 1 3 3
    1 2 6
    */

[Croatia OI 2002 Final Exam - First day]: http://poj.org/searchproblem?field=source&key=Croatia+OI+2002+Final+Exam+-+First+day
[20160815161157996]: /images/20220718/3e21bc05ba3c4cdcb8bbec1e7e77a4c7.png