poj1273 Drainage Ditches 基础网络流

#include <iostream>
#include <queue>
using namespace std;
int G[300][300];
int Prev[300]; //路径上每个节点的前驱节点
bool Visited[300]; 
int n,m;  //m是顶点数目，顶点编号从1开始 1是源，m是汇, n是 边数

int Augment()
{  
    int v;
    int i;
    deque<int> q;
    memset(Prev,0,sizeof(Prev));
    memset(Visited,0,sizeof(Visited));
    Prev[1] = 0;
    Visited[1] = 1;
    q.push_back(1);
    bool bFindPath = false;
  //用bfs寻找一条源到汇的可行路径 
    while (!q.empty())
    {
        v = q.front();
        q.pop_front();
        for (i = 1; i <= m; i++)
        {
            if (G[v][i] > 0 && Visited[i] == 0)
            {   
                //必须是依然有容量的边，才可以走
                Prev[i] = v;
                Visited[i] = 1;
                if( i == m )
                {
                    bFindPath = true;
                    q.clear();
                    break;
                }
                else 
                    q.push_back(i);
            } 
        }
    } 
            
    if (!bFindPath)
        return 0;
    int nMinFlow = 999999999;
    v = m;  //寻找源到汇路径上容量最小的边，其容量就是此次增 加的总流量
    while( Prev[v] )
    {   
        nMinFlow = min( nMinFlow,G[Prev[v]][v]); 
        v = Prev[v];
    } 
    //沿此路径添加反向边，同时修改路径上每条边的容量
    v = m; 
    while( Prev[v] )
    {  
        G[Prev[v]][v] -= nMinFlow;
        G[v][Prev[v]] += nMinFlow; 
        v = Prev[v];
    } 
    return nMinFlow;
} 
            
int main() 
{
    while (cin >> n >> m) 
    {   
        //m是顶点数目，顶点编号从1开始
        int i,j,k;
        int s,e,c;
        memset( G,0,sizeof(G));
        for( i = 0;i < n;i ++ )
        {    
            cin >> s >> e >> c;  
            G[s][e] += c; //两点之间可能有多条边
        }   
        int MaxFlow = 0; 
        int aug; 
        while( aug = Augment() )
            MaxFlow += aug;  
        cout << MaxFlow << endl; 
    }  
    return 0; 
}