FZU 1046 Tempter of the Bone
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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.
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.
'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.
4 4 5S.X...X...XD....3 4 5S.X...X....D0 0 0
NOYES
#include<iostream>#include<algorithm>#include<cstdio>using namespace std;int n,m,t;int si,sj,di,dj;int dir[4][2]={0,1,0,-1,1,0,-1,0};char map[9][9];int flag;void dfs(int si,int sj,int time){if(si>n||si<1||sj>m||sj<1)return;if(time==t&&si==di&&sj==dj)flag=1;if(flag)return ;if(((si+sj)%2==(di+dj)%2)&&((t-time)%2==1))return;if(((si+sj)%2!=(di+dj)%2)&&((t-time)%2==0))return;for(int i=0;i<4;i++){if(map[si+dir[i][0]][sj+dir[i][1]]!='X'){map[si+dir[i][0]][sj+dir[i][1]]='X';dfs(si+dir[i][0],sj+dir[i][1],time+1);map[si+dir[i][0]][sj+dir[i][1]]='.';}}return;}int main(){while(cin>>n>>m>>t){if(n==0&&m==0&&t==0)break;int wall=0;//判断墙的数目;for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){cin>>map[i][j];if(map[i][j]=='S'){si=i;sj=j;}if(map[i][j]=='D'){di=i;dj=j;}if(map[i][j]=='X')wall++;}}map[si][sj]='X';flag=0;dfs(si,sj,0);if(flag)cout<<"YES"<<endl; else cout<<"NO"<<endl; } return 0;}
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