POJ 1511 Invitation Cards

POJ 1511 Invitation Cards

构造反向图，分别求单源最短路，然后求和即可。
以下是我的代码：

#include < iostream >
#include < queue >
#include < cstdio >
#include < cstring >
using   namespace  std;
const   int  kMaxn( 1000007 );
const   int  kMaxm( 1000007 );
const   long   long  kInf( 0x7f7f7f7f7f7f7f7f );
struct  Edge
{
     int  v,w;
};

void  SPFA( int  n, int  first[kMaxn], int  next[kMaxm],Edge e[kMaxm], long   long  d[kMaxn])
{
    queue < int >  q;
     bool  inq[kMaxn];
    memset(inq, false , sizeof (inq));
     for ( int  i = 1 ;i <= n;i ++ )
        d[i] = kInf;
    d[ 1 ] = 0 ;
    q.push( 1 );
    inq[ 1 ] = 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 (d[v] > d[u] + w)
            {
                d[v] = d[u] + w;
                 if ( ! inq[v])
                {
                    q.push(v);
                    inq[v] = true ;
                }
            }
        }
    }
}

int  n,m;
int  cnt,first[kMaxn],next[kMaxm];Edge e[kMaxm];
int  cnt2,first2[kMaxn],next2[kMaxm];Edge e2[kMaxm];
long   long  d[kMaxn],d2[kMaxn];

int  main()
{
     int  T;
    scanf( " %d " , & T);
     while (T -- )
    {
        cnt = cnt2 = 0 ;
        memset(first, - 1 , sizeof (first));
        memset(first2, - 1 , sizeof (first2));

        scanf( " %d%d " , & n, & m);
         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;

            cnt2 ++ ;
            e2[cnt2].v = u;e2[cnt2].w = w;
            next2[cnt2] = first2[v];
            first2[v] = cnt2;
        }

        SPFA(n,first,next,e,d);
        SPFA(n,first2,next2,e2,d2);

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

        cout << ans << endl;
    }

     return   0 ;
}