POJ 1459 Power Network

经典最大流问题，需要将题中的多源点多汇点最大流问题转化为单源单汇。

源点s指向power station，边的权重为相应的power station的p_max；

_{consumer都指向汇点t，边权为相应的consumer的cmax；}

采用Edmonds-Karps耗时有点大，借助scanf才能勉强在1s内跑完！待学习了压入与重标记算法后重写代码。

#include <iostream>#include <vector>#include <cstdio>#include <queue>using namespace std;int n, np, nc, m;int s, t;vector<vector<int> > links;vector<int> path, flow;int MaxFlow(){    int maxFlow = 0;    while( true )    {path.assign(n + 2, 0);path[s] = s;flow.assign(n + 2, 0);flow[s] = 0xffff;queue<int> q;q.push( s );while( !q.empty() ){    int u = q.front();    if( u == t )break;    q.pop();    for(int j = 0; j < n + 2; ++j)    {if( !flow[j] && links[u][j] ){    path[j] = u;    flow[j] = min(flow[u], links[u][j]);    q.push( j );}    }}if( q.empty() ){    break;}maxFlow += flow[t];for(int v = t; v != s; v = path[v]){    int u = path[v];    links[u][v] -= flow[t];    links[v][u] += flow[t];}    }    return maxFlow;}int main(){    while( cin >> n >> np >> nc >> m )    {links.clear();links.resize( n + 2 );for(int i = 0; i < links.size(); ++i)    links[i].assign(n+2, 0);s = n ;t = n + 1;int u, v, w; char ch;for(int i = 0; i < m; ++i){    scanf(" (%d,%d)%d", &u, &v, &w);    links[u][v] = w;}for(int i = 0; i < np; ++i){    scanf(" (%d)%d", &v, &w);    links[s][v] = w;}for(int i = 0; i < nc; ++i){    scanf(" (%d)%d", &u, &w);    links[u][t] = w;}cout << MaxFlow() << endl;    }    return 0;}