dfs/poj 3009 Curling 2.0

 1 #include<cstdio>
 2 using namespace std;
 3 
 4 const int dx[4]={0,1,0,-1};
 5 const int dy[4]={1,0,-1,0};
 6 
 7 int m,n,ans,ex,ey,sx,sy;
 8 int a[22][22];
 9 
10 int cmin(int a,int b) {return a<b?a:b;}
11 
12 bool pd(int xx,int yy)
13 {
14     if (xx>=1 && xx<=n && yy>=1 && yy<=m) return true;
15     return false;
16 }
17 
18 void dfs(int x,int y,int step)
19 {
20     if (step>10 || step>=ans ) return ;
21 
22     for (int i=0;i<4;i++)
23     {
24         int xx=x+dx[i],yy=y+dy[i];
25         if (a[xx][yy]==1) continue;
26         while (1)
27         {
28             if (!pd(xx,yy)) break;
29             if (a[xx][yy]==1)
30             {
31                 a[xx][yy]=0;
32                 dfs(xx-dx[i],yy-dy[i],step+1);
33                 a[xx][yy]=1;
34                 break;
35             }
36             else if (a[xx][yy]==3)
37             {
38                 ans=cmin(ans,step+1);
39                 return;
40             }
41             xx+=dx[i];yy+=dy[i];
42         }
43     }
44     return;
45 }
46 
47 int main()
48 {
49     scanf("%d%d",&m,&n);
50     while (m!=0)
51     {
52         for (int i=1;i<=n;i++)
53             for (int j=1;j<=m;j++)
54             {
55                 scanf("%d",&a[i][j]);
56                 if (a[i][j]==2){sx=i;sy=j;a[i][j]=0;}
57             }
58 
59         ans=11;
60         dfs(sx,sy,0);
61         if (ans>10) printf("-1\n");
62         else printf("%d\n",ans);
63 
64         scanf("%d%d",&m,&n);
65     }
66     return 0;
67 }