POJ 1273 Drainage Ditches

POJ 1273 Drainage Ditches

求网络流的最大流。
注意：可能出现重边！比如第一次c[u][v]=w1，又有c[u][v]=w2，那么从u到v的容量应该为w1+w2！
以下是我的代码：

#include < queue >
#include < algorithm >
#include < cstdio >
#include < cstring >
#define  kS 1
#define  kT n
using   namespace  std;
const   int  kMaxn( 207 );

int  n,m,maxflow,c[kMaxn][kMaxn],f[kMaxn][kMaxn];

int  main()
{
     while (scanf( " %d%d " , & m, & n) == 2 )
    {
        memset(c, 0 ,kMaxn * kMaxn * sizeof ( int ));
        memset(f, 0 ,kMaxn * kMaxn * sizeof ( int ));
         for ( int  i = 1 ;i <= m;i ++ )
        {
             int  u,v,w;
            scanf( " %d%d%d " , & u, & v, & w);
            c[u][v] += w;
        }

        maxflow = 0 ;
         while ( true )
        {
             int  d[kMaxn],p[kMaxn];
            queue < int >  q;
            memset(d, 0 ,kMaxn * sizeof ( int ));
            d[kS] = 0x7f7f7f7f ;
            q.push(kS);
             while ( ! q.empty())
            {
                 int  u(q.front());q.pop();
                 for ( int  v = 1 ;v <= n;v ++ )
                     if ( ! d[v]  &&  c[u][v] > f[u][v])
                    {
                        p[v] = u;q.push(v);
                        d[v] = min(d[u],c[u][v] - f[u][v]);
                    }
            }
             if (d[kT] == 0 )
                 break ;
            maxflow += d[kT];
             for ( int  u = kT;u != kS;u = p[u])
            {
                f[p[u]][u] += d[kT];
                f[u][p[u]] -= d[kT];
            }
        }

        printf( " %d\n " ,maxflow);
    }

     return   0 ;
}