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Power Network(最大流基础_增广路算法：多源多汇，自建总源点和总汇点)

2024-07-18 07:42:55   219人阅读


Power NetworkCrawling in process...Crawling failedTime Limit:2000MS    Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
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Description

A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied with an amount s(u) >= 0 of power, may produce an amount 0 <= p(u) <= pmax(u) of power, may consume an amount 0 <= c(u) <= min(s(u),cmax(u)) of power, and may deliver an amount d(u)=s(u)+p(u)-c(u) of power. The following restrictions apply: c(u)=0 for any power station, p(u)=0 for any consumer, and p(u)=c(u)=0 for any dispatcher. There is at most one power transport line (u,v) from a node u to a node v in the net; it transports an amount 0 <= l(u,v) <= lmax(u,v) of power delivered by u to v. Let Con=Σuc(u) be the power consumed in the net. The problem is to compute the maximum value of Con.

An example is in figure 1. The label x/y of power station u shows that p(u)=x and pmax(u)=y. The label x/y of consumer u shows that c(u)=x and cmax(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and lmax(u,v)=y. The power consumed is Con=6. Notice that there are other possible states of the network but the value of Con cannot exceed 6.

Input

There are several data sets in the input. Each data set encodes a power network. It starts with four integers: 0 <= n <= 100 (nodes), 0 <= np <= n (power stations), 0 <= nc <= n (consumers), and 0 <= m <= n^2 (power transport lines). Follow m data triplets (u,v)z, where u and v are node identifiers (starting from 0) and 0 <= z <= 1000 is the value of lmax(u,v). Follow np doublets (u)z, where u is the identifier of a power station and 0 <= z <= 10000 is the value of pmax(u). The data set ends with nc doublets (u)z, where u is the identifier of a consumer and 0 <= z <= 10000 is the value of cmax(u). All input numbers are integers. Except the (u,v)z triplets and the (u)z doublets, which do not contain white spaces, white spaces can occur freely in input. Input data terminate with an end of file and are correct.

Output

For each data set from the input, the program prints on the standard output the maximum amount of power that can be consumed in the corresponding network. Each result has an integral value and is printed from the beginning of a separate line.

Sample Input

2 1 1 2 (0,1)20 (1,0)10 (0)15 (1)20
7 2 3 13 (0,0)1 (0,1)2 (0,2)5 (1,0)1 (1,2)8 (2,3)1 (2,4)7
         (3,5)2 (3,6)5 (4,2)7 (4,3)5 (4,5)1 (6,0)5
         (0)5 (1)2 (3)2 (4)1 (5)4

Sample Output

15
6

Hint

The sample input contains two data sets. The first data set encodes a network with 2 nodes, power station 0 with pmax(0)=15 and consumer 1 with cmax(1)=20, and 2 power transport lines with lmax(0,1)=20 and lmax(1,0)=10. The maximum value of Con is 15. The second data set encodes the network from figure 1.

ps：坑我的一道题啊，题目看了好长时间才明白，那么多数据直接吓哭我了，输入的时候也调试了好久，以后再也不在输入的时候随便打空格了。怎么会有这样的呢！输出按ctrl+z

题意：有n个结点，np个发电站，nc个消费者，m个电力运输线。接下去是n条边的信息(u,v)cost，cost表示边(u,v)的最大流量；a个发电站的信息(u)cost，cost表示发电站u能提供的最大流量；b个用户的信息(v)cost，cost表示每个用户v能接受的最大流量。
思路：在图中添加1个源点S和汇点T，将S和每个发电站相连，边的权值是发电站能提供的最大流量；将每个用户和T相连，边的权值是每个用户能接受的最大流量。从而转化成了一般的最大网络流问题，然后求解。（自己语言总结不好，觉得这个能看懂，就贴一下）


#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <iostream>
#include <queue>
using namespace std;
#define inf  9999999999
int flow[210][210];
int maxflow[210],father[210],vis[210];
int max_flow;
int m,i;
void EK(int s,int e)
{
    queue<int >q;
    int u,v;
    max_flow=0;
    while(1)
    {
        memset(maxflow,0,sizeof(maxflow));
        memset(vis,0,sizeof(vis));
        maxflow[s]=inf;
        q.push(s);
        while(!q.empty())
        {
            u=q.front();
            q.pop();
            for(v=s;v<=e;v++)
            {
                if(!vis[v]&&flow[u][v]>0)
                {
                    vis[v]=1;
                    father[v]=u;
                    q.push(v);
                    maxflow[v]=min(maxflow[u],flow[u][v]);
                }
            }
            if(maxflow[e]>0)
            {
                while(!q.empty())
                    q.pop();
                break;
            }
        }
        if(maxflow[e]==0)
            break;
        for(i=e;i!=s;i=father[i])
        {
            flow[father[i]][i]-=maxflow[e];
            flow[i][father[i]]+=maxflow[e];
        }
        max_flow+=maxflow[e];
    }
}
int main()
{
    int n,np,nc,m;
	int i,a,b,c;
	char ch;
	while(~scanf("%d%d%d%d ",&n,&np,&nc,&m))//输入注意后面的空格
	{
		memset(flow,0,sizeof(flow));
		for(i=1;i<=m;i++)
		{
			scanf("(%d,%d)",&a,&b);
			scanf("%d ",&c);//注意后面的空格
			flow[a+1][b+1]=c;
		}
		for(i=1;i<=np;i++)//建立总源点
		{
			scanf("(%d)%d ",&a,&b);//注意后面空格
			flow[0][a+1]=b;
		}
		for(i=1;i<=nc;i++)//建立总汇点
		{
            scanf("(%d)%d ",&a,&b);//注意后面的空格
			flow[a+1][n+1]=b;
		}
		EK(0,n+1);
		printf("%d\n",max_flow);
	}
	return 0;
}




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