最小步数NYOJ

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <queue>
 4 using namespace std;
 5 
 6 int d[9][9];  //标记此点是否访问并记录距离
 7 int maze[9][9]={
 8     1,1,1,1,1,1,1,1,1,
 9     1,0,0,1,0,0,1,0,1,
10     1,0,0,1,1,0,0,0,1,
11     1,0,1,0,1,1,0,1,1,
12     1,0,0,0,0,1,0,0,1,
13     1,1,0,1,0,1,0,0,1,
14     1,1,0,1,0,1,0,0,1,
15     1,1,0,1,0,0,0,0,1,
16     1,1,1,1,1,1,1,1,1};
17 const int INF=1<<10;
18 typedef pair<int ,int> P;
19 int n,sx,sy,gx,gy;
20 int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};   
21 
22 int bfs()
23 {
24     memset(d,0,sizeof(d));
25     queue<P> que;
26     for(int i=0;i<9;i++)
27         for(int j=0;j<9;j++)
28             d[i][j]=INF;
29     que.push(P(sx,sy));
30     d[sx][sy]=0;
31     
32     while(!que.empty())
33     {
34         P p=que.front();
35         que.pop();
36         if(p.first==gx&&p.second==gy)
37             break;
38         
39         for(int i=0;i<4;i++)
40         {
41             int nx=p.first+dx[i],ny=p.second+dy[i];
42             
43             if(0<=nx&&nx<=8&&0<=ny&&ny<=8&&maze[nx][ny]==0&&d[nx][ny]==INF)  //当前位置是否可以加入
44             {
45                 que.push(P(nx,ny));
46                 d[nx][ny]=d[p.first][p.second]+1;
47             }
48         }
49     }
50     return d[gx][gy];
51 }
52 
53 int main()
54 {
55     scanf("%d",&n);
56     while(n--)
57     {
58         scanf("%d%d%d%d",&sx,&sy,&gx,&gy);
59         printf("%d\n",bfs());
60     }
61     return 0;
62 }