【HDU 1010】Tempter of the Bone（DFS）

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

题目大意

给你一个n*m的图，问你doggie能否在T时间刚好从S走到D（doggie每走一步需要1s的时间），如果可以输出YES，否则输出NO。

思路

简单的DFS，每跑一步就将记录的步数+1，如果走到D之后步数（时间）刚好等于T，则可行。

代码

#include <iostream>#include <stdio.h>#include <string.h>#include <stdlib.h>#include <math.h>using namespace std;const int BLANK=8;const int LINE=8;char str[BLANK][LINE];int posX,posY,n,m,t,flag,numcount,stax,stay;;int direction[2][4]={{1,-1,0,0},{0,0,1,-1}};int chess[BLANK][LINE];void solve(int x,int y,int numcount){    int i,desX,desY;    if(x==posX&&y==posY)    {        if(t==numcount)            flag=1;        return;    }    if(numcount>t) return;    if(str[x][y]!='X')    {        for(i=0;i<4;i++)        {            desX=x+direction[0][i];            desY=y+direction[1][i];            if(str[desX][desY]!='X'&&desX>=1&&desX<=n&&desY>=1&&desY<=m&&!chess[desX][desY])            {                chess[desX][desY]=1;                solve(desX,desY,numcount+1);                chess[desX][desY]=0;                if(flag)                    return;            }          }    }}int main(){    while(scanf("%d %d %d",&n,&m,&t)!=EOF)    {        if(n==0&&m==0&&t==0) break;        for(int i=1;i<=n;i++)        {            getchar();            for(int j=1;j<=m;j++)            {                scanf("%c",&str[i][j]);                if(str[i][j]=='S') {stax=i;stay=j;}                if(str[i][j]=='D') {posX=i;posY=j;}            }        }        getchar();        memset(chess,0,sizeof(chess));        if(abs(stax-posX)+abs(stay-posY)>t||(stax+posX+stay+posY+t)%2==1)        {            printf("NO\n");            continue;        }        chess[stax][stay]=1;        flag=0;        numcount=0;        solve(stax,stay,numcount);        if(flag==1) printf("YES\n");        else printf("NO\n");    }    return 0;}