SDAU 搜索专题 18 Tempter of the Bone

1：问题描述
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

2：大致题意

小狗进入了一个陷阱，x是墙，s是小狗的初始位置，d是门的位置。
不过门只有在特定的时间可以打开，而且小狗在一个地方只能待1个单位时间。问小狗能不能出去。

3：思路
这个用BFS好像就解决不了了，要用DFS。用DFS的时候要根据奇偶进行剪枝。
结束条件有：
1 越界。
2 墙。
3 剩余步数减去最短的步数为奇数。
4 到达指定位置并且门正好开放。要记录下来。
如果没有结束，就利用方向数组来遍历4个方向，遍历的时候要标记已经走过的路。并且已经走过的路加1.递归之后要恢复现场。

4：感想

不太会用DFS。〒_〒。让我再研究研究。。。

5：ac代码

#include<iostream>#include<cstdio>#include<stdio.h>#include<cstring>#include<cmath>using namespace std;int n,m,t,ex,ey,tt,falg;char mapp[10][10];int tox[5]={0,0,1,-1};int toy[5]={1,-1,0,0};void dfs(int x,int y,int num){    if(x<0||y<0||x>=n||y>=m)        return;    if(mapp[x][y]=='X')        return ;    if(x==ex&&y==ey&&num==t)        falg=1;    if(falg)    {        tt=num;        return ;    }    int tem=t-fabs(x-ex)-fabs(y-ey)-num;    if(tem<0||((tem%2)==1))       return ;    for(int i=0;i<4;i++)    {        mapp[x][y]='X';        dfs(x+tox[i],y+toy[i],num+1);        mapp[x][y]='.';    }}int main(){   //freopen("r.txt","r",stdin);    int i,j,sx,sy;    while(cin>>n>>m>>t)    {        if((t+n+m)==0) break;        int wall=0;        for(i=0;i<n;i++)        {            scanf("%s",mapp[i]);            for(j=0;j<m;j++)            {                if(mapp[i][j]=='S')                {                    sx=i;                    sy=j;                }                if(mapp[i][j]=='D')                {                    ex=i;                    ey=j;                }                if(mapp[i][j]=='X')                    wall++;            }        }        if((n*m-wall-1)<t)        {            cout<<"NO"<<endl;            continue;        }        falg=0;        dfs(sx,sy,0);        if(falg)            cout<<"YES"<<endl;        else            cout<<"NO"<<endl;    }}