Contest 8A The Maze runner

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本文介绍了一种求解迷宫最短路径的算法实现，通过深度优先搜索（DFS）来寻找从起点到终点的最短路径。文章详细解释了输入格式、迷宫的定义、移动约束及输出要求，并提供了一个C语言实现的示例代码。

Problem A: The Maze runner

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 468 Solved: 256
[ Submit][ Status][ Web Board]

Description

Given a maze, find a shortest path from start to exit.

Input

Input consists serveral test cases.

First line of the input contains number of test case T.

For each test case the first line contains two integers N , M ( 1 <= N, M <= 100 ).

Each of the following N lines contain M characters. Each character means a cell of the map.

Here is the definition for chracter.

Constraint:

A map consists of 4 kinds of characters:
‘S’ means the start point;
‘E’ means the exit point;
‘.’ means the opened-block that you can pass;
‘#’ means the closed-block that you cannot pass;

Requirements:
It is ONLY allowed to move by one step vertically or horizontally( up, down , left or right) to the next block;
You MUST NOT get out of the map;
Return -1 if given arguments can not satisfy the above requirements or you cannot find a way from ‘S’ to ‘E’.

Output

For each test case print one line containing shortest path. If there exists no path from start to goal, print -1.

Sample Input

4
4 4
####
#S##
#.E#
####
6 5
#####
#.S.#
#.#.#
#..##
##.E#
#####
4 4
####
#S##
##E#
####
6 8
########
#.....##
#.##...#
#..#.#.#
#S#..E.#
########

Sample Output

HINT

#include<stdio.h>
char maze[20][20];
int x,y;
int min=-1;
int move[4][2]={{0,1},{1,0},{-1,0},{0,-1}};
void DFS(int a,int b,int step)
{
 int i,tx,ty;
 for(i=0;i<4;i++){
  tx=a+move[i][0];
  ty=b+move[i][1];
  if(tx<0||tx>=x||ty<0||ty>=y||maze[tx][ty]=='#')
   continue;
  if(maze[tx][ty]=='E'&&step+1<min){
      min=step+1;
      return;
 }
  if(maze[tx][ty]!='E')
      maze[tx][ty]='#';
  DFS(tx,ty,step+1);
  if(maze[tx][ty]!='E')
         maze[tx][ty]='-';
 }
}
int main()
{
 int T;
 scanf("%d",&T);
 while(T-->0){
  int i,j;
  scanf("%d%d",&x,&y);
  for(i=0;i<x;i++)
   scanf("%s",maze[i]);
  for(i=0;i<x;i++)
   for(j=0;j<y;j++)
    if(maze[i][j]=='S'){
     min=99;
     maze[i][j]='#';
     DFS(i,j,0);
     break;
    }
    if(min==99)
     min=-1;
  printf("%d\n",min);
 }
 return 0;
}