FYOI 2010 信使

FYOI 2010 信使

构图+SPFA求最短路即可，只不过代码很长。
通过这道题目我发现数组模拟链表比动态内存分配构建链表快得多！
以下是我的代码：

#include < stdio.h >
#include < string .h >
#define  maxn 307
#define  INF 510000007
FILE  * fin = fopen( " message.in " , " r " ), * fout = fopen( " message.out " , " w " );
typedef  struct
{
     long  v,w,next;
}edge;
typedef  struct
{
     long  count,front,rear,item[maxn * maxn * 10 ];
}queue;
const   long  xd[] = { - 1 , - 1 , 0 , 1 , 1 , 1 , 0 , - 1 },
           yd[] = { 0 , 1 , 1 , 1 , 0 , - 1 , - 1 , - 1 };
long  n,m,x,y,k,c[ 3 ];
long  r[maxn][maxn],a[maxn][maxn];
long  first[maxn * maxn],count;
edge e[maxn * maxn * 10 ];
queue q;
long  d[maxn * maxn];
bool  inq[maxn * maxn];

void  Clear()
{
    q.count = 0 ;q.front = 0 ;q.rear =- 1 ;
}
void  Push( long  x)
{
    q.count ++ ;q.rear ++ ;q.item[q.rear] = x;
}
void  Pop( long   & x)
{
    q.count -- ;x = q.item[q.front];q.front ++ ;
}
bool  Empty()
{
     return  (q.count == 0 );
}
long  f( long  _x, long  _y)
{
     return  (_x - 1 ) * m + _y;
}
void  input()
{
    fscanf(fin, " %ld%ld%ld%ld%ld%ld%ld\n " , & n, & m, & x, & y, & k, & c[ 1 ], & c[ 2 ]);
    c[ 0 ] = 0 ;
     for ( long  i = 1 ;i <= n;i ++ )
    {
        for ( long  j = 1 ;j <= m;j ++ )
       {
           char  t;
          fscanf(fin, " %c " , & t);
           switch (t)
          {
              case   ' F ' :r[i][j] = 0 ; break ;
              case   ' H ' :r[i][j] = 1 ; break ;
              case   ' B ' :r[i][j] = 2 ;
          }
       }
       fscanf(fin, " \n " );
    }
     for ( long  i = 1 ;i <= n;i ++ )
       for ( long  j = 1 ;j <= m;j ++ )
        fscanf(fin, " %ld " , & a[i][j]);
}
void  build()
{
    count = 0 ;
     for ( long  i = 1 ;i <= n;i ++ )
     for ( long  j = 1 ;j <= m;j ++ )
    {
       first[f(i,j)] = 0 ;
        for ( long  k = 0 ;k < 8 ;k ++ )
        if (i + xd[k] >= 1 && i + xd[k] <= n && j + yd[k] >= 1 && j + yd[k] <= m)
       {
          count ++ ;
          e[count].v = f(i + xd[k],j + yd[k]);
          e[count].w = a[i + xd[k]][j + yd[k]];
           if (r[i][j] != r[i + xd[k]][j + yd[k]]) e[count].w += c[r[i + xd[k]][j + yd[k]]];
          e[count].next = first[f(i,j)];
          first[f(i,j)] = count;
       }
    }
}
void  spfa()
{
     long  s = f(x,y),t,p;
     for ( long  i = 1 ;i <= f(n,m);i ++ )
      d[i] = (i == s ? a[x][y] + c[r[x][y]]:INF);
    memset(inq, false , sizeof (inq));
    Clear();
    Push(s);inq[s] = true ;
     while ( ! Empty())
    {
       Pop(t);inq[t] = false ;
        for (p = first[t];p;p = e[p].next)
          if (d[t] + e[p].w < d[e[p].v])
         {
            d[e[p].v] = d[t] + e[p].w;
             if ( ! inq[e[p].v])
            {
               Push(e[p].v);inq[e[p].v] = true ;
            }
         }
    }
}
void  output()
{
     long  ans = 0 ;
     for ( long  i = 1 ;i <= k;i ++ )
    {
        long  begin,end;
       fscanf(fin, " %ld%ld " , & begin, & end);
       ans += d[f(begin,end)];
    }
    fprintf(fout, " %ld\n " ,ans);
}
void  close()
{
    fclose(fin);fclose(fout);
}
int  main()
{
    input();
    build();
    spfa();
    output();
    close();
return   0 ;
}