杭电-1010 Tempter of the Bone（深搜）

Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 106936    Accepted Submission(s): 29087

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

 

Author

ZHANG, Zheng

 

Source

ZJCPC2004

 

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#include<stdio.h>#include<string.h>int vis[10][10];//用来记录这个点是否被访问过char mat[10][10];//用来记录这个点int ok,t,m,n;//这里的ok是用来记录判断的，如果深搜确认是可以逃出陷阱，ok被标记为1void visit(int i,int j,int cur)//（i，j）为当前访问的坐标，cur指访问点（i，j）是第cur步，刚开始cur为0{    if(cur>t||ok==1) return;//如果步数超过了T，理所当然要退出啦。或者如果之前的方案确定可以逃出陷阱，理所当然要退出啦。    else if(cur==t)  //如果当前步数为T，那么要判断当前是否到了门口{        if(mat[i][j]=='D')//是的话，令ok=1            ok=1;        return;    }    else if(mat[i][j]=='.'||mat[i][j]=='S'){  //如果当前访问的格子合法（不要漏了“S”，不然第一次的访问将失效）        vis[i][j]=1;  //做上标记        if(i+1>=0&&i+1<n&&j>=0&&j<m&&vis[i+1][j]==0)  visit(i+1,j,cur+1);  //判断下一个格子是否在定义域内且是否被标记过，是，访问        if(i-1>=0&&i-1<n&&j>=0&&j<m&&vis[i-1][j]==0)  visit(i-1,j,cur+1);        if(i>=0&&i<n&&j+1>=0&&j+1<m&&vis[i][j+1]==0)  visit(i,j+1,cur+1);        if(i>=0&&i<n&&j-1>=0&&j-1<m&&vis[i][j-1]==0)  visit(i,j-1,cur+1);        vis[i][j]=0;//千万记得消除标记，这步要读者深思    }}int main(){    while(scanf("%d%d%d",&n,&m,&t)!=EOF){        if(n==0||m==0||t==0)            break;        ok=0;//记得初始化        for(int i=0;i<n;i++){         scanf("%s",mat[i]);  //输入图        }    memset(vis,0,sizeof(vis));//初始化，将vis【】【】数组全部赋值为0；    int a,b;    for(int i=0;i<n;i++){        for(int j=0;j<m;j++){            if(mat[i][j]=='S')            a=i,b=j;//记录S的坐标        }    }    visit(a,b,0);//从S开始访问    if(ok==0)        printf("NO\n");    else if(ok==1)        printf("YES\n");}    return 0;}