# HDU-1010 Tempter of the Bone Deep search template problem (DPS) Report

Source: Internet
Author: User

Tempter of the BoneTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 88587 Accepted Submission (s): 24116

Problem DescriptionThe 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 is a trap, and he tried desperately to get out of this maze.

The maze is a rectangle with sizes N by M. There is a door in the maze. At the beginning, the door is 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 in 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 on one block for more than one second, nor could he move into a visited block. Can the poor doggie survive? Please help him.

Inputthe input consists of multiple test cases. The first line of all test case contains three integers n, m, and T (1 < N, m < 7; 0 < T <), which denote The sizes of the maze and the time at which the door would open, respectively. The next N lines give the maze layout, with all 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 is a not-to-be processed.

Outputfor each test case, print on 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`

Authorzhang, ZhengTest Instructions:

Give you a maze of n rows M-Columns, x represents the Wall, "." On behalf of the road, s for the starting point, D for the end, ask can just in k seconds from the beginning to reach the end.

Exercises

Template Dfs, no pruning required. Open a two-dimensional array to go up and down, open a oount record steps, if count>k, has not come to the direct break.

Reference code:

`#include <cmath> #include <iostream>using namespace Std;char str[10][10];int map[10][10],v[4][2]={0,1,0,-    1,1,0,-1,0};int n,m,k,i,j,flag,ax,ay,bx,by;void dfs (int x,int y,int count) {int mx,my;        if (bx==x&&by==y) {if (k==count) flag=1;    Return    } if (count>=k) return;            if (str[x][y]!= ' x ') {for (int i=0;i<4;i++) {mx=x+v[i][0];            MY=Y+V[I][1]; if (mx>=1&&mx<=n&&my>=1&&my<=m&&!map[mx][my]&&str[mx][my]!=                ' X ') {map[mx][my]=1;                DFS (MX,MY,COUNT+1);                map[mx][my]=0;            if (flag) return; }}}}int Main () {while (scanf ("%d%d%d", &n,&m,&k)!=eof&& (n+m+k)) {memset (ma        P,0,sizeof (map));        int count;            for (i=1;i<=n;i++) {GetChar ();          for (j=1;j<=m;j++)  {scanf ("%c", &str[i][j]);                    if (str[i][j]== ' S ') {ax=i;                Ay=j;                    } if (str[i][j]== ' D ') {bx=i;                By=j;        }}} GetChar (); if (ABS (BX-AX) +abs (by-ay) >k| |        (ax+bx+ay+by+k)%2==1) printf ("no\n");            else {flag=0;            count=0;            Map[ax][ay]=1;            DFS (Ax,ay,count);            if (flag) printf ("yes\n");        else printf ("no\n");  }} return 0;}`

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HDU-1010 Tempter of the Bone Deep search template problem (DPS) Report

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