POJ 3268 Silver Cow Party

POJ 3268 Silver Cow Party

正向图反向图各求一次单源最短路。
以下是我的代码：

#include < iostream >
#include < queue >
#include < algorithm >
#include < cstdio >
#include < cstring >
using   namespace  std;
const   int  kMaxn( 1007 );
const   int  kMaxm( 100007 );
const   long   long  kInf(0x7f7f7f7f7f7f7f7fll);
struct  Edge
{
     int  v,w;
};

int  n,m,x;
int  cnt,first[kMaxn],next[kMaxm];Edge e[kMaxm];
int  first2[kMaxn],next2[kMaxm];Edge e2[kMaxm];
long   long  d1[kMaxn],d2[kMaxn];

void  SPFA( int  first[kMaxn], int  next[kMaxn],Edge e[kMaxm], int  start, long   long  dist[kMaxn])
{
    queue < int >  q;
     bool  inq[kMaxn] = { false };
     for ( int  i = 1 ;i <= n;i ++ )
        dist[i] = (i == start ? 0 :kInf);
    q.push(start);
    inq[start] = true ;
     while ( ! q.empty())
    {
         int  u(q.front());q.pop();
        inq[u] = false ;
         for ( int  i = first[u];i !=- 1 ;i = next[i])
        {
             int  v(e[i].v),w(e[i].w);
             if (dist[v] > dist[u] + w)
            {
                dist[v] = dist[u] + w;
                 if ( ! inq[v])
                {
                    q.push(v);
                    inq[v] = true ;
                }
            }
        }
    }
}

int  main()
{
     while (scanf( " %d%d%d " , & n, & m, & x) == 3 )
    {
        cnt = 0 ;
        memset(first, - 1 , sizeof (first));
        memset(first2, - 1 , sizeof (first));
         for ( int  i = 1 ;i <= m;i ++ )
        {
             int  u,v,w;
            scanf( " %d%d%d " , & u, & v, & w);
            cnt ++ ;
            e[cnt].v = v;e[cnt].w = w;
            next[cnt] = first[u];
            first[u] = cnt;
            e2[cnt].v = u;e2[cnt].w = w;
            next2[cnt] = first2[v];
            first2[v] = cnt;
        }

        SPFA(first,next,e,x,d1);
        SPFA(first2,next2,e2,x,d2);

         long   long  ans( 0 );
         for ( int  i = 1 ;i <= n;i ++ )
            ans = max(ans,d1[i] + d2[i]);

        cout << ans << endl;
    }

     return   0 ;
}