Power Network(poj1459 网络最大流 EK法）

Power Network

Time Limit: 2000MS Memory Limit: 32768KTotal Submissions: 29098 Accepted: 15093

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)207 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

156

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

#include<iostream>#include<algorithm>#include<fstream>#include<math.h>#include<algorithm>#include<stack>#include<queue>using namespace std;const int inf = 1 << 29;const int MAXN = 310;int n, np, nc, m;int map[MAXN][MAXN];//邻接矩阵存图bool vis[MAXN];//访问标记int pre[MAXN];//记录前一个点void init()//初始化{int len = n + 2;//因为要加上超级源点和超级汇点for (int i = 0; i < len; i++){for (int j = 0; j < len; j++){map[i][j] = 0;}}}int getmin(int a, int b){return a < b ? a : b;}int getroute(int s, int e)//BFS求源点到汇点的路径{int len = n + 2;int temp;for (int i = 0; i < len; i++){vis[i] = false;pre[i] = -1;}bool haveroute = false;queue<int> que;que.push(s);vis[s] = true;while (!haveroute && !que.empty()){temp = que.front();que.pop();for (int i = 0; i < len; i++){if (map[temp][i] && !vis[i]){vis[i] = true;pre[i] = temp;if (i == e){haveroute = true;break;}que.push(i);}}}if (!haveroute)return false;return true;}int getMaxflow(int s, int e)//求最大流{int t;int result = 0;while (getroute(s, e)){int minflow = inf;t = e;while (pre[t] != -1)//寻找路线中最小的线路{minflow = getmin(minflow, map[pre[t]][t]);//遍历该路径求最小容量/最大流量t = pre[t];}t = e;while (pre[t] != -1)//增广路{map[pre[t]][t] -= minflow;//减去已用的容量map[t][pre[t]] += minflow;//增加反向边t = pre[t];}result += minflow;}return result;}int main(){char rubbish;//去掉糟糕的括号int first, next;int value;while (cin >> n >> np >> nc >> m){init();int start = n;//超级源点int end = n + 1;//超级汇点for (int i = 0; i < m; i++){cin >> rubbish;cin >> first;cin >> rubbish;cin >> next;cin >> rubbish;cin >> value;if (first == next)continue;map[first][next] = value;}for (int i = 0; i < np; i++){cin >> rubbish;cin >> first;cin >> rubbish;cin >> value;map[start][first] = value;//超级源点与供电点相连}for (int i = 0; i < nc; i++){cin >> rubbish;cin >> first;cin >> rubbish;cin >> value;map[first][end] = value;//超级汇点与消费点相连}cout << getMaxflow(start, end) << endl;}return 0;}

转载地址：http://blog.csdn.net/u012171516/article/details/49202239
Edmonds-Karp算法:http://www.cnblogs.com/jackge/archive/2013/05/05/3061893.html