hdu1548 奇怪的电梯 dfs dijkstra bfs都可以，在此奉上dfs

题目链接：http://icpc.njust.edu.cn/Problem/Hdu/5706/

简单的规定深度进行搜索，代码如下：

 #include<bits/stdc++.h>
 using namespace std;
 typedef unsigned int ui;
 typedef long long ll;
 typedef unsigned long long ull;
 #define pf printf
 #define mem(a,b) memset(a,b,sizeof(a))
 #define prime1 1e9+7
 #define prime2 1e9+9
 #define pi 3.14159265
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define scand(x) scanf("%llf",&x)
 #define f(i,a,b) for(int i=a;i<=b;i++)
 #define scan(a) scanf("%d",&a)
 #define dbg(args) cout<<#args<<":"<<args<<endl;
 #define inf 0x3f3f3f3f
 #define maxn 1005
 #define maxm 2000010
 int n,m,t;
 char Map[maxn][maxn];
 int ans1=,ans2=;
 char s[][]={{"cat"},{"girl"}};
 int dir[][]={,,-,,,,,-};
 void dfs(int x,int y,int flag,int cur)
 {
     int xx,yy;
     if(flag==)
     {
         if(cur==)
         {
             ans1++;
             return;
         }
         f(i,,)
         {
              xx=x+dir[i][];
              yy=y+dir[i][];
             if(xx>=&&xx<n&&yy>=&&yy<m&&Map[xx][yy]==s[flag][cur])
             {
                 dfs(xx,yy,,cur+);
             }
         }
     }
     else
     {
         if(cur==)
         {
             ans2++;
             return;
         }
         f(i,,)
         {
              xx=x+dir[i][];
              yy=y+dir[i][];
             if(xx>=&&xx<n&&yy>=&&yy<m&&Map[xx][yy]==s[flag][cur])
             {
                 dfs(xx,yy,,cur+);
             }
         }
     }
 }
 int main()
 {
     //freopen("input.txt","r",stdin);
     //freopen("output.txt","w",stdout);
     std::ios::sync_with_stdio(false);
     scan(t);
     while(t--)
     {
         ans1=ans2=;
         scan(n);
         scan(m);
         f(i,,n-)scanf("%s",Map[i]);
         f(i,,n-)
             f(j,,m-)
             {
                 if(Map[i][j]=='g')
                 {
                     dfs(i,j,,);
                 }
                 if(Map[i][j]=='c')
                 {
                     dfs(i,j,,);
                 }
             }
             pf("%d %d\n",ans1,ans2);
     }
  }