题目描述
Consider an N x N (1 <= N <= 100) square field composed of 1
by 1 tiles. Some of these tiles are impassible by cows and are marked with an 'x' in this 5 by 5 field that is challenging to navigate:
. . B x .
. x x A .
. . . x .
. x . . .
. . x . . Bessie finds herself in one such field at location A and wants to move to location B in order to lick the salt block there. Slow, lumbering creatures like cows do not like to turn and, of course, may only move parallel to the edges of the square field. For a given field, determine the minimum number of ninety degree turns in any path from A to B. The path may begin and end with Bessie facing in any direction. Bessie knows she can get to the salt lick.
N*N(1<=N<=100)方格中，’x’表示不能行走的格子，’.’表示可以行走的格子。卡门很胖，故而不好转弯。现在要从A点走到B点，请问最少要转90度弯几次？
输入输出格式
输入格式：
第一行一个整数N,下面N行，每行N个字符，只出现字符：’.’，’x’，’A’，’B’，表示上面所说的矩阵格子，每个字符后有一个空格。
【数据规模】
2<=N<=100
输出格式：
一个整数：最少转弯次数。如果不能到达，输出-1。
输入输出样例
3 . x A . . . B x .
2
说明
【注释】
只可以上下左右四个方向行走，并且不能走出这些格子之外。开始和结束时的方向可以任意。
1//我用的DFS，他们说dfs做不出来，但我做出来了QWQ
#include<bits/stdc++.h> 2 using namespace std; 3 int read() 4 { 5 int ret=0,ok=1;char ch=getchar(); 6 while(ch<'0'||ch>'9') {if(ch=='-')ok=-1;ch=getchar();} 7 for(;ch>='0'&&ch<='9';ch=getchar()) ret=ret*10+ch-'0'; 8 return ret*ok; 9 } 10 int n; 11 char a[105][105]; 12 int ans[105][105]; 13 bool vis[105][105],sea[105][105]; 14 int bx[]={-1,1,0,0}; 15 int by[]={0,0,-1,1}; 16 inline void dfs(int sx,int sy) 17 { 18 sea[sx][sy]=true; 19 if(sx>n || sy>n || sy<=0 || sx<=0) 20 return; 21 vis[sx][sy]=1; 22 for(int i=sx+1;i<=n;i++) 23 { 24 if(a[i][sy]=='x') 25 break; 26 if(ans[i][sy]>ans[sx][sy]+1) 27 { 28 ans[i][sy]=ans[sx][sy]+1; 29 dfs(i,sy); 30 } 31 vis[i][sy]=true; 32 } 33 for(int i=sx-1;i>=1;i--) 34 { 35 if(a[i][sy]=='x') 36 break; 37 if(ans[i][sy]>ans[sx][sy]+1) 38 { 39 ans[i][sy]=ans[sx][sy]+1; 40 dfs(i,sy); 41 } 42 vis[i][sy]=true; 43 } 44 for(int i=sy+1;i<=n;i++) 45 { 46 if(a[sx][i]=='x') 47 break; 48 if(ans[sx][i]>ans[sx][sy]+1) 49 { 50 ans[sx][i]=ans[sx][sy]+1; 51 dfs(sx,i); 52 } 53 vis[sx][i]=true; 54 } 55 for(int i=sy-1;i>=1;i--) 56 { 57 if(a[sx][i]=='x') 58 break; 59 if(ans[sx][i]>ans[sx][sy]+1) 60 { 61 ans[sx][i]=ans[sx][sy]+1; 62 dfs(sx,i); 63 } 64 vis[sx][i]=true; 65 } 66 for(int i=1;i<=4;i++) 67 { 68 int nx=sx+bx[i]; 69 int ny=sy+by[i]; 70 if(nx>n || ny>n || ny<=0 || nx<=0) 71 continue; 72 if(vis[nx][ny] && !sea[nx][ny]) 73 { 74 sea[nx][ny]=1; 75 dfs(nx,ny); 76 } 77 } 78 } 79 int main() 80 { 81 int sx,sy; 82 int lx,ly; 83 for(int i=1;i<=100;i++) 84 for(int j=1;j<=100;j++) 85 ans[i][j]=1234567; //这一步赋值很关键！之前用memset赋0x7f，会错一个点 86 n=read(); 87 for(int i=1;i<=n;i++) 88 for(int j=1;j<=n;j++) 89 { 90 cin>>a[i][j]; 91 if(a[i][j]=='A') 92 { 93 sx=i; 94 sy=j; 95 ans[i][j]=-1; 96 } 97 if(a[i][j]=='B') 98 { 99 lx=i; 100 ly=j; 101 } 102 } 103 dfs(sx,sy); 104 if(ans[lx][ly]==1234567) 105 { 106 cout<<-1<<endl; 107 return 0; 108 } 109 cout<<ans[lx][ly]<<endl; 110 return 0; 111 }