Tempter of the Bone

The doggie found a bone in an ancient maze, which fascinated him a lot. However, when he picked it up, the maze began to shake, and the doggie could feel the ground sinking. He realized that the bone was a trap, and he tried desperately to get out of this maze. 

The maze was a rectangle with sizes N by M. There was a door in the maze. At the beginning, the door was closed and it would open at the T-th second for a short period of time (less than 1 second). Therefore the doggie had to arrive at the door on exactly the T-th second. In every second, he could move one block to one of the upper, lower, left and right neighboring blocks. Once he entered a block, the ground of this block would start to sink and disappear in the next second. He could not stay at one block for more than one second, nor could he move into a visited block. Can the poor doggie survive? Please help him. 

InputThe input consists of multiple test cases. The first line of each test case contains three integers N, M, and T (1 < N, M < 7; 0 < T < 50), which denote the sizes of the maze and the time at which the door will open, respectively. The next N lines give the maze layout, with each line containing M characters. A character is one of the following: 

'X': a block of wall, which the doggie cannot enter; 
'S': the start point of the doggie; 
'D': the Door; or 
'.': an empty block. 

The input is terminated with three 0's. This test case is not to be processed. 
OutputFor each test case, print in one line "YES" if the doggie can survive, or "NO" otherwise. 
Sample Input

4 4 5
S.X.
..X.
..XD
....
3 4 5
S.X.
..X.
...D
0 0 0

Sample Output

NO
YES

思路：除了基础的DFS和常规剪枝外，多了一个剪枝：奇偶剪枝

奇偶剪枝：代码中  剩余可用步数（时间）-最短到达终点的步数   结果如果为奇数则剪去

因为走不同于最短路径，相当于绕路，而绕路必定与最短路径（这个最短路径中也包括终点，从起点走到终点，走截然不同的路径，也可以理解成绕路）的所用步数差值为偶数，这样才能回到正确的到达路径上。

AC代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
int en,em;
int n,m,t;
char map[8][8];
int vis[8][8];
int flag=0;
int x[4]={0,0,-1,1};
int y[4]={1,-1,0,0};
void dfs(int a,int b,int ans){
    vis[a][b]=1;
    if(flag)return ;
    if(ans==t&&map[a][b]=='D'){flag=1;return ;}
    if(ans>t)return ;
     int temp=t-ans-abs(a-en)-abs(b-em);//此为奇偶剪枝
     if(temp&1)return ;

   for(int i=0;i<4;i++){
    int tempa=a+x[i];
    int tempb=b+y[i];
    if(tempa>=0&&tempa<n&&tempb>=0&&tempb<m&&vis[tempa][tempb]==0&&map[tempa][tempb]!='X'){
            vis[tempa][tempb]=1;
        dfs(tempa,tempb,ans+1);
    vis[tempa][tempb]=0;
    }
   }

}
int main(){
    int sn,sm;
    while(cin>>n>>m>>t&&n&&m&&t){
            flag=0;
    memset(vis,0,sizeof(vis));
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                cin>>map[i][j];
                if(map[i][j]=='S'){
                    sn=i;sm=j;
                }
                if(map[i][j]=='D'){
                    en=i;em=j;
                }
            }
        }
        dfs(sn,sm,0);
        if(flag)printf("YES\n");
        else printf("NO\n");

    }
return 0;
}