HDU 1072 优先队列的使用

思路很简单，一开始没有用优先队列，WA了几次后来改了，想一想走迷宫停顿时应该用优先队列。if(u.btime==1)  continue;此句不能省略当时间为1时，这一步无法出去，只有重新来过

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  1 #include<cstdio>
  2 #include<cstdlib>
  3 #include<iostream>
  4 #include<cstring>
  5 #include<queue>
  6 
  7 using namespace std;
  8 int N,M,map[10][10];
  9 int dx[] = {0,0,-1,1};
 10 int dy[] = {1,-1,0,0};
 11 typedef struct
 12 {
 13    int x,y;
 14    int step;
 15    int btime;
 16 }Point;
 17 Point P[100];  
 18 
 19 int True(Point P)
 20 {
 21     if(P.x<N&&P.x>=0&&P.y>=0&&P.y<M)
 22         return 1;
 23      return 0;
 24 }
 25 
 26 priority_queue<Point> q;
 27 bool operator<(Point a, Point b)
 28 {
 29      return a.step > b.step;
 30  }       
 31  
 32 int main()
 33 {
 34     int i,j,ncases;
 35     int sx,sy,tx,ty,ok;
 36     Point start,target;
 37     
 38     scanf("%d",&ncases);
 39     while( ncases-- )
 40     {
 41       scanf("%d%d",&N,&M);        
 42       for(i=0; i<N; i++)
 43        for(j=0; j<M; j++) 
 44        {
 45            scanf("%d",&map[i][j]);
 46        }
 47        for(i=0; i<N; i++)
 48         for(j=0; j<M; j++)
 49         {
 50            if(map[i][j] == 2)
 51            {
 52               start.x = i;
 53               start.y = j;
 54            }
 55            if(map[i][j] == 3)             
 56            {
 57              target.x = i;
 58              target.y = j;
 59            }            
 60         }
 61         start.step = 0;
 62         ok = 0;
 63         start.btime = 6;
 64         q.push(start);
 65         while( !q.empty() )
 66         {
 67            Point u = q.top();
 68             q.pop();   
 69            if(u.x==target.x && u.y==target.y)
 70            {  
 71                 ok = 1;
 72                 printf("%d\n",u.step);
 73                 break;
 74            }                
 75             if(u.btime == 1) continue;
 76             for(i=0; i<4; i++)        
 77             {
 78               Point v;
 79               v.x = u.x + dx[i];
 80               v.y = u.y + dy[i];
 81               if(True(v) && map[v.x][v.y])
 82               {            
 83                  v.btime = u.btime-1;  
 84                  v.step = u.step + 1;                     
 85                  if(map[v.x][v.y]==4)
 86                  {
 87                     v.btime = 6;
 88                     map[v.x][v.y] = 0;
 89                  } 
 90                  q.push(v);   
 91               }
 92             }
 93          }
 94         while( !q.empty())
 95            q.pop();    
 96        if(!ok)
 97          printf("-1\n");
 98      }        
 99     system("pause");
100     return 0;
101 } 
102