Tempter of the Bone
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 58461 Accepted Submission(s): 15902
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.
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.
'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 5
S.X.
..X.
..XD
....
3 4 5
S.X.
..X.
...D
0 0 0
Sample Output
NO
YES
Author
ZHANG, Zheng
Source
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1 //62MS 232K 1306 B G++ 2 /* 3 4 题意： 5 给出一个图，求点S到点D是否存在步数为K的路径 6 7 DFS： 8 不剪枝的话复杂度约为 4^49 必TLE 9 此处最重要的剪枝是奇偶剪枝，所谓奇偶剪枝， 10 就是判断剩余步数与目标的最短路径是否符合要求，不符合则先排除 11 从而优化。 12 13 */ 14 #include<stdio.h> 15 #include<string.h> 16 #define N 15 17 char g[N][N]; 18 int vis[N][N]; 19 int mov[4][2]={0,1,1,0,0,-1,-1,0}; 20 int n,m,k,flag; 21 int ex,ey; 22 int Abs(int a) 23 { 24 return a>0?a:-a; 25 } 26 void dfs(int x,int y,int cnt) 27 { 28 if(flag || cnt>k) return; 29 if(g[x][y]=='D' && cnt==k) flag=1; 30 int temp=(k-cnt)-Abs(x-ex)-Abs(y-ey); 31 if(temp<0 || temp&1) return; 32 for(int i=0;i<4;i++){ 33 int x0=x+mov[i][0]; 34 int y0=y+mov[i][1]; 35 if(x0>=0 && x0<n && y0>=0 && y0<m && !vis[x0][y0] && g[x0][y0]!='X'){ 36 vis[x0][y0]=1; 37 dfs(x0,y0,cnt+1); 38 vis[x0][y0]=0; 39 } 40 } 41 } 42 int main(void) 43 { 44 int sx,sy; 45 while(scanf("%d%d%d",&n,&m,&k),n+m+k) 46 { 47 int cnt=1; 48 for(int i=0;i<n;i++){ 49 scanf("%s",g[i]); 50 for(int j=0;j<m;j++) 51 if(g[i][j]=='S') 52 sx=i,sy=j; 53 else if(g[i][j]=='.') cnt++; 54 else if(g[i][j]=='D') 55 ex=i,ey=j; 56 } 57 if(cnt<k){ 58 puts("NO"); 59 continue; 60 } 61 memset(vis,0,sizeof(vis)); 62 vis[sx][sy]=1; 63 flag=0; 64 dfs(sx,sy,0); 65 if(flag) puts("YES"); 66 else puts("NO"); 67 } 68 return 0; 69 }