HDU 1010 Tempter of the Bone

Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 62925    Accepted Submission(s): 17223

Problem Description

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.

 

Input

The 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.

 

Output

For each test case, print in one line "YES" if the doggie can survive, or "NO" otherwise.

 

Sample Input

4 4 5S.X...X...XD....3 4 5S.X...X....D0 0 0

 

Sample Output

NOYES

 

题意：狗在迷宫中在S处开始跑，D处为门，门只在第T秒开，也就是狗必须恰好在第T秒跑到D处才能得救，其中迷宫中的X表示墙不能走，. 表示可以走。问狗能否逃离。

直接DFS超时，求教别人后加了一个奇偶剪枝就过了。

关于奇偶剪枝，点击此处：奇偶剪枝

#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>const int MAX = 9;int visit[MAX][MAX];char map[MAX][MAX];int sx,sy,dx,dy,n,m,t;bool mark;void dfs(int x,int y,int count){if(x<0 || y<0 || x>=n || y>=m)return;if(count>t || mark)return;if(visit[x][y]==1)return;if(map[x][y]=='X')return;if(map[x][y]=='D' && count==t){mark = true;return;}if(((abs(x-dx)+abs(y-dy))&1) != ((t-count)&1))return;//奇偶剪枝visit[x][y] = 1;dfs(x-1,y,count+1);dfs(x+1,y,count+1);dfs(x,y+1,count+1);dfs(x,y-1,count+1);visit[x][y] = 0;}void init(){int i,j;for(i=0;i<MAX;++i){for(j=0;j<MAX;++j){visit[i][j] = 0;}}}int main(){int i,j,count;//freopen("in.txt","r",stdin);    while(scanf("%d %d %d%*c",&n,&m,&t)!=EOF){if(n==0 && m==0 && t==0)break;init();for(i=0;i<n;++i){for(j=0;j<m;++j){map[i][j]=getchar();if(map[i][j]=='S'){sx = i;sy = j;}else if(map[i][j]=='D'){dx = i;dy = j;}}getchar();}count = 0;mark = false;dfs(sx,sy,count);if(mark){printf("YES\n");}else{printf("NO\n");}    }    return 0;}