POJ 3169 Layout

查分约束好题，关键在于转化！

 

#include < iostream >
using   namespace  std;
#define  MAX 20020
#define  INF 999999
struct  Node{
     int  u,v;
     int  val;
}node[MAX];
int  nv,ne,start;
int  dist[MAX];
int  bellman_DC()
{
     int  i,j,tmp,tag;
     for (i = 1 ;i <= nv; ++ i)
        dist[i]  =  INF;
    dist[start]  =   0 ;
     for (i = 1 ;i < nv; ++ i)
    {
        tag  =   0 ;
         for (j = 1 ;j <= ne; ++ j)
        {
            tmp  =  dist[node[j].u] + node[j].val;
             if (tmp < dist[node[j].v])
            {
                dist[node[j].v]  =  tmp;
                tag  =   1 ;
            }
        }
         if (tag  ==   0 )
             return   0 ;
    }
     for (i = 1 ;i <= ne; ++ i)
         if (dist[node[i].v] > dist[node[i].u] + node[i].val)
             return   1 ;
     return   0 ;
}
int  main()
{
     int  i,j,a,b,flag,m1,m2,val,index;
     while (scanf( " %d%d%d " , & nv, & m1, & m2) != EOF)
    {
        index  =   0 ;
        ne  =  m1 + m2;
         for (i = 1 ;i <= m1; ++ i)
        {
             ++ index;
            scanf( " %d%d%d " , & a, & b, & val);
            node[index].u  =  a;
            node[index].v  =  b;
            node[index].val  =  val;
        }
         for (i = 1 ;i <= m2; ++ i)
        {
             ++ index;
            scanf( " %d%d%d " , & a, & b, & val);
            node[index].u  =  b;
            node[index].v  =  a;
            node[index].val  =   - val;
        }
        start  =   1 ;
        flag  =  bellman_DC();
         if (flag  ==   1 )
            printf( " -1\n " );
         else   if (dist[nv] == INF)
            printf( " -2\n " );
         else  
            printf( " %d\n " ,dist[nv]);
    }
     return   0 ;
}