最少步数----深搜

 1  
 2 #include <stdio.h>
 3 int tu[9][9]={
 4 1,1,1,1,1,1,1,1,1,
 5 1,0,0,1,0,0,1,0,1,
 6 1,0,0,1,1,0,0,0,1,
 7 1,0,1,0,1,1,0,1,1,
 8 1,0,0,0,0,1,0,0,1,
 9 1,1,0,1,0,1,0,0,1,
10 1,1,0,1,0,1,0,0,1,
11 1,1,0,1,0,0,0,0,1,
12 1,1,1,1,1,1,1,1,1,
13 };
14 int bs[9][9]={0};
15 int min=32;
16 
17 int DFS(int a,int b,int c,int d,int count)/*起始位置（a,b），终点位置（c,d）,步数count 初值为 0 */
18 {
19     if(tu[a][b]==1) return count;
20     else
21     {
22         if(a==c&&b==d)
23         {
24             if(min>count) min=count;
25         }
26         bs[a][b]=1;
27 
28         if((!tu[a-1][b])&&(!bs[a-1][b]))/*上边*/
29             count=DFS(a-1,b,c,d,count+1);
30         if((!tu[a+1][b])&&(!bs[a+1][b]))/*下边*/
31             count=DFS(a+1,b,c,d,count+1);
32         if((!tu[a][b-1])&&(!bs[a][b-1]))/*左边*/
33             count=DFS(a,b-1,c,d,count+1);
34         if((!tu[a][b+1])&&(!bs[a][b+1]))/*右边*/
35             count=DFS(a,b+1,c,d,count+1);
36 
37         bs[a][b]=0;
38 
39         return count-1;
40     }
41 }
42 
43 int main(void)
44 {
45     int a,b,c,d,n,count;
46     scanf("%d",&n);
47     while(n--)
48     {
49         count=0;
50         min=30;
51         scanf("%d%d%d%d",&a,&b,&c,&d);
52         if(tu[a][b]==0 && tu[c][d]==0)
53         {
54             DFS(a,b,c,d,count);
55             printf("%d\n",min);
56         }
57     }
58     return 0;
59 }
60