hdu1010



temper of bone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 73261    Accepted Submission(s): 20105

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

 

Sample Output

NO
YES


源代码
#include<iostream>
#include<string.h>
using namespace std;

const int temp=1;
int load[6][6];                     //构造的0、1矩阵，便于找点
int n,m,r,t;
char map[6][6];
void search_road(int i,int j,int time)
{
    if(time>t||r==1) return;
    else if(time==t)
        if(map[i][j]=='D'){
            r=1;return;
        }

    if(map[i][j]=='S'||map[i][j]=='.'){
        load[i][j]=temp;
        if(i+1>-1&&i+1<n&&j>-1&&j<m&&load[i+1][j]==0) search_road(i+1,j,time+1);
        if(i-1>-1&&i-1<n&&j>-1&&j<m&&load[i-1][j]==0) search_road(i-1,j,time+1);
        if(i>-1&&i<n&&j-1>-1&&j-1<m&&load[i][j-1]==0) search_road(i,j-1,time+1);
        if(i>-1&&i<n&&j+1>-1&&j+1<m&&load[i][j+1]==0) search_road(i,j+1,time+1);
        load[i][j]=0;
    }
}
int main()
{
    while(cin>>n>>m>>t){
  if(n==0&&m==0&&t==0) exit(0);
  for(int i=0;i<n;i++) 
   cin>>map[i];                   //输入地图格式
  int a,b;
        for(i=0;i<n;i++)
   for(int j=0;j<m;j++)
    if(map[i][j]=='S'){
                a=i,b=j;          //找起始点
        }
  memset(load,0,sizeof(load));    //赋0
  r=0;
  search_road(a,b,0);
  if(r==1)
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
}

这道题是一道很中规中矩的迷宫类型的题，主要难点在于函数递归，作为上手题难度还可以。刚编写完运行的时候发现输出结果始终是NO，检查了几遍才发现递归部分最后忘记把load归零了，这会导致后来检测周围点是否被标记时出错。