POJ 1459 Power Network

經典最大流問題，需要將題中的多源點多匯點最大流問題轉化爲單源單匯。

源點s指向power station，邊的權重爲相應的power station的p_max；

_{consumer都指向匯點t，邊權爲相應的consumer的cmax；}

採用Edmonds-Karps耗時有點大，藉助scanf才能勉強在1s內跑完！待學習了壓入與重標記算法後重寫代碼。

#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;
}