NYOJ58——最小步数，简单的BFS应用

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #include<queue>
 6 
 7 using namespace std;
 8 
 9 int dx[4] = {1,-1, 0, 0};
10 int dy[4] = {0, 0, 1,-1};
11 int visit[9][9];
12 int map[9][9] = {{1,1,1,1,1,1,1,1,1},{1,0,0,1,0,0,1,0,1},{1,0,0,1,1,0,0,0,1},{1,0,1,0,1,1,0,1,1},
13 {1,0,0,0,0,1,0,0,1},{1,1,0,1,0,1,0,0,1},{1,1,0,1,0,1,0,0,1},{1,1,0,1,0,0,0,0,1},{1,1,1,1,1,1,1,1,1}};    
14     
15 typedef struct
16 {
17   int x, y;
18   int step;      
19 }Point;
20 
21 int True(Point V)
22 {
23     if(V.x>=0&&V.x<=8&&V.y>=0&&V.y<=8)
24        return 1;
25     return 0;
26 }
27 
28 queue<Point>q;   
29 int main()
30 {
31     int i, j, ncases, flag;
32     Point Start, Target;
33     
34     scanf("%d", &ncases);
35     getchar();
36     while( ncases-- )
37     {
38        scanf("%d %d %d %d",&Start.x,&Start.y,&Target.x,&Target.y);
39        getchar();  
40        memset(visit, 0, sizeof(visit)); 
41        visit[Start.x][Start.y] = 1;
42        Start.step = 0;
43        q.push(Start);
44        while( !q.empty() )
45        {
46           Point u = q.front();
47           q.pop();     
48           if(u.x==Target.x&&u.y==Target.y&&True(u))//到达目的地打印退出。
49           {
50               printf("%d\n", u.step);
51               break;
52           }
53           for(i=0; i<4; i++)
54           {
55              Point v;
56              v.x = u.x + dx[i]; 
57              v.y = u.y + dy[i];
58              v.step = u.step + 1;
59              if(map[v.x][v.y]!=1 && !visit[v.x][v.y] && True(v))
60              {
61                 visit[v.x][v.y] = 1;
62                 q.push(v);
63              } 
64           } 
65        }   
66        while(!q.empty())
67        {      
68           q.pop();
69        }
70      }
71      return 0;
72 }
73