題目鏈接：http://poj.org/problem?id=1459

Power Network

Time Limit:?2000MS	?	Memory Limit:?32768K
Total Submissions:?27074	?	Accepted:?14066

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.

Source

Southeastern Europe 2003 題意：給幾個發電站，給幾個消耗站，再給幾個轉發點。發電站只發電，消耗站只消耗電，轉發點只是轉發電，再給各個傳送線的傳電能力。問你消耗站能獲得的最多電是多少。 分析：加一個超級源點，和一個超級匯點。 這個題之前做過。記錄一下EK的思想吧。 EK: 廣搜每一層節點，每次都記錄增廣路上的最小流量，匯點的最小流量為0時，說明沒有增廣路了。根據匯點的最小流量，更新流。 有一點要注意的是，搜索第一層后，記得標記，再往下一層搜，一直搜到匯點為止，但是我這里好像是沒有標記，其實標記了，node[i]沒有標記為0. 輸入有空格。 #include <stdio.h> #include <string.h> #include <queue> #include <iostream>using namespace std;#define MAX 120 #define INF 0x3f3f3f3fint n,np,nc,m; int cap[MAX][MAX];int main() {//freopen("input.txt","r",stdin);int from,to,value;while(scanf("%d%d%d%d",&n,&np,&nc,&m)!=EOF){memset(cap,0,sizeof(cap));///讀取輸電線的數據while(m--){scanf(" (%d,%d)%d",&from,&to,&value);cap[from][to]=value;}///讀取發電站數據while(np--){scanf(" (%d)%d",&from,&value);cap[n][from]=value;}///讀取消費者數據while(nc--){scanf(" (%d)%d",&from,&value);cap[from][n+1]=value;}int ans = 0;queue<int> Q;int flow[MAX][MAX]; ///剩余網絡int node[MAX]; ///最小流int pre[MAX]; ///增廣路徑 memset(flow,0,sizeof(flow));while(true){Q.push(n);memset(node,0,sizeof(node));node[n] = INF;int u;while(!Q.empty()){u = Q.front();Q.pop();for(int i=0; i<=n+1; i++){if(!node[i]&&cap[u][i]>flow[u][i]){Q.push(i);node[i] = min(node[u],cap[u][i]-flow[u][i]);pre[i] = u;}}}if(node[n+1]==0)break;for(u=n+1; u!=n; u=pre[u]){flow[pre[u]][u] +=node[n+1];flow[u][pre[u]] -= node[n+1];}ans+=node[n+1];}printf("%d\n",ans);}return 0; }

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轉載于:https://www.cnblogs.com/TreeDream/p/5752109.html

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