Power Network - poj 1459 (最大流 Edmonds-Karp算法)

 

Time Limit: 2000MS		Memory Limit: 32768K
Total Submissions: 24788		Accepted: 12922

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) <= p_max(u) of power, may consume an amount 0 <= c(u) <= min(s(u),c_max(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) <= l_max(u,v) of power delivered by u to v. Let Con=Σ_uc(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 p_max(u)=y. The label x/y of consumer u shows that c(u)=x and c_max(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and l_max(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 l_max(u,v). Follow np doublets (u)z, where u is the identifier of a power station and 0 <= z <= 10000 is the value of p_max(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 c_max(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.

 

中文说一下

多组数据，每组先是有四个数据 N,np,nc,M 其中N表示节点数（包括用电用户，中继站，发电站的总数）np表示发电站的个数，nc表示，用电户的个数，M表示电线的个数

然后再输出M组数据表示电线，每根电线用（u,v）z表示表示从u到v最多可以传送的电量为z

再输出np个发电站，每个发电站形式是（u）z表示发电站u可以发电最多z

最后输出nc个用电户，每个用电户的形式是（u）z，表示用电户u最多会用z的电

这题用最大流方法可以求解。这题使用求Ford-Fulkerson方法求最大流，使用BFS的算法称为Edmonds-Karp算法。

#include <iostream>
#include<stdlib.h>
#include<stdio.h>
#include<string.h>
#include<queue>
#define INF 100000
using namespace std;
int map[103][103],path[103],flow[103],start,end;
queue<int> q;
int BFS(){
    while(!q.empty())q.pop();
    memset(path,-1,sizeof(path));
    q.push(start);
    path[start]=0;
    flow[start]=INF;
    while(!q.empty()){
        int v=q.front();
        q.pop();
        if(v==end)
            break;
        for(int i=1;i<=end;i++){
            if(i!=start&&map[v][i]&&path[i]==-1){
                flow[i]=flow[v]<map[v][i]?flow[v]:map[v][i];
                q.push(i);
                path[i]=v;
            }
        }
    }
    if(path[end]==-1) return -1;
    else
        return flow[end];
}
int Edmonds_Karp(){
    int max_flow,step,now,pre;
    max_flow=0;
    while(1){
        step=BFS();
        if(step==-1)
            break;
        max_flow+=step;
        now=end;
        while(now!=start){
            pre=path[now];
            map[pre][now]-=step;
            map[now][pre]+=step;
            now=pre;
        }
    }
    return max_flow;
}
int main() {
    int n,np,nc,m;

    while(~scanf("%d %d %d %d",&n,&np,&nc,&m)){
        memset(map,0,sizeof(map));
        getchar();
        for(int i=0;i<m;i++){
            int u,v,f;
            scanf(" (%d,%d)%d ",&u,&v,&f);
            map[u+2][v+2]=f;

        }
        for(int i=0;i<np;i++){
            int s,f;
            scanf(" (%d)%d ",&s,&f);
            map[1][s+2]=f;

        }
        for(int i=0;i<nc;i++){
            int s,f;
            scanf(" (%d)%d ",&s,&f);
            map[s+2][n+2]=f;
        }
        start=1;
        end=n+2;
        int result=Edmonds_Karp();
        printf("%d\n",result);
    }

    return 0;
}