hdu 1010 Tempter of the Bone 简单带回溯的dfs+奇偶剪枝

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1010

题意：

有一只小狗需要从N*M迷宫里逃离，出发点标记为'S','X'为不可走，'D'为出口，出口是一扇门，只有在T时刻才打开。问小狗能否逃离迷宫？

T (1 < N, M < 7; 0 < T < 50)

简单带回溯的dfs，很久以前做过，时隔两年，为了复习算法又遇此题。结果还超时了，一脸懵逼，然后果断看以前的代码(我好菜)，加了一个奇偶剪枝，AC。

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<vector>using namespace std;#define all(x) (x).begin(), (x).end()#define for0(a, n) for (int (a) = 0; (a) < (n); (a)++)#define for1(a, n) for (int (a) = 1; (a) <= (n); (a)++)#define mes(a,x,s)  memset(a,x,(s)*sizeof a[0])#define mem(a,x)  memset(a,x,sizeof a)#define ysk(x)  (1<<(x))typedef long long ll;//const int INF =  ;const int maxn= 7 ;int n,m,T,endX,endY;char a[maxn+2][maxn+2];bool vis[maxn+2][maxn+2];int dir[4][2]={ {-1,0},{0,+1},{+1,0},{0,-1}  };bool in(int x,int y){return 0<=x&&x<=n-1&&0<=y&&y<=m-1;}bool dfs(int x,int y,int time){if(abs(x-endX)+abs(y-endY)>T-time || (T-time-abs(x-endX)+abs(y-endY))%2   )  return false;for0(j,4){int tx=x+dir[j][0],ty=y+dir[j][1];if(!in(tx,ty)||vis[tx][ty]||a[tx][ty]=='X') continue;if(a[tx][ty]=='D'&&time+1==T)return true;vis[tx][ty]=1;if(dfs(tx,ty,time+1))  return true;vis[tx][ty]=0;}return false;}int main(){int startX,startY;while(~scanf("%d%d%d",&n,&m,&T)&& n+m+T!=0){for0(i,n)  for0(j,m)  {scanf(" %c",&a[i][j]);if(a[i][j]=='S'){startX=i,startY=j;}else if(a[i][j]=='D'){endX=i,endY=j;}}mem(vis,0);vis[startX][startY]=1;puts(dfs(startX,startY,0)?"YES":"NO");}return 0;}