Wikioi 1215 maze

 

 

1215 maze 6 people are recommended to add questions to favorites

 

Description Description

In the N * n maze, "#" is the wall, "." is the path, "S" is the start point, and "e" is the end point. A total of four directions can be taken. Go to the (n-1, n-1) "E") position in the upper-left corner (0, 0) "S"). You can use the general rule to output yes. If not, no is output.

Input description Input description

The first input behavior is an integer m, indicating the number of maze. Then, the first behavior of each maze data is an integer n (n ≤ 16), which indicates the length of the sides of the maze. The next n rows contain N characters in each line and are separated by spaces.

Output description Output description

The output contains m rows. If the maze corresponding to each row can go, yes is output; otherwise, no is output.

Sample Input Sample Input

1  7  s...##.  .#.....  .......  ..#....  ..#...#  ###...#  ......e

Sample output Sample output

YES

 

Simply put, this question is simply explained by the DFS and then expanded to the surrounding area.

#include<stdio.h>#include<math.h>char a[1001][1001],ans=0;int directx[5]={0,0,0,-1,1},vis[1001][1001];int directy[5]={0,-1,1,0,0};void dfs(int i,int j,int n){    if(i<=0||i>n||j<=0||j>n)return ;    if(vis[i][j]==1||a[i][j]==‘#‘)return ;    if(a[i][j]==‘e‘){ans=a[i][j];    printf("YES");        }    vis[i][j]=1;    for(int x=1;x<=4;x++)    dfs(i+directx[x],j+directy[x],n) ;}int main(){    int b,i,z,o,j,n;    scanf("%d",&z);   for(o=1;o<=z;o++)   {       scanf("%d\n",&n);       for(i=1;i<=n;i++)       for(j=1;j<=n;j++)       {           if(j<n)           scanf("%c",&a[i][j]);           else scanf("%c\n",&a[i][j]);       }       dfs(1,1,n);    if(ans==0)printf("NO");       printf("\n");   }      return 0;}