ZOJ Problem Set - 2110 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.
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
不减少递归条件就会TLE，使劲了各种手段、查题解，修改递归函数，调试了半天才过了。。。
#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
char maze[10][10];
int si,sj,ei,ej;
int n,m,t;
int flag;
int dir[4][2]={{0,-1},{0,1},{-1,0},{1,0}};
int abs(int num)
{
if(num<0)return -1*num;
else return num;
}
void dfs(int i,int j,int ct)
{
if(i<=0||i>n||j<=0||j>m)
{
return ;
}
else if(ct>t)return;
else if(i==ei&&j==ej&&ct==t)
{
//cout<<"========in 'D'======"<<endl;
//cout<<"i: "<<i<<" j: "<<j<<" ct: "<<ct<<endl;
//cout<<maze[i][j]<<endl;
//cout<<"========in 'D'======"<<endl;
flag=true;
return;
}
int temp=(t-ct)-abs(i-ei)-abs(j-ej);//减少递归分支，以免超时
//cout<<"temp: "<<temp<<endl;
if(temp<0||temp%2)return;
for(int k=0;k<4;k++)
{
//cout<<"k: "<<k<<endl;
if(maze[i+dir[k][0]][j+dir[k][1]]=='.')//检查周围的点是否是一个点 {
//cout<<"i: "<<i<<" j: "<<j<<" ct: "<<ct<<endl;
maze[i+dir[k][0]][j+dir[k][1]]='X';
dfs(i+dir[k][0],j+dir[k][1],ct+1);
maze[i+dir[k][0]][j+dir[k][1]]='.';
if(flag)return;
}
}
}
int main()
{
while(cin>>n>>m>>t)
{
if(n==0&&m==0&&t==0)break;
memset(maze,0,sizeof(maze));
//memset(visit,0,sizeof(maze));
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
cin>>maze[i][j];
if(maze[i][j]=='S')
{
si=i;sj=j;
maze[i][j]='X';
}
else if(maze[i][j]=='D')
{
ei=i;ej=j;
maze[i][j]='.';
}
}
}
flag=false;
dfs(si,sj,0);
if(flag)
{
cout<<"YES"<<endl;
}
else cout<<"NO"<<endl;
}
}
