The Maze runner（BFS）

最新推荐文章于 2022-10-31 21:23:50 发布

原创最新推荐文章于 2022-10-31 21:23:50 发布 · 660 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#bfs #graph

基础专栏收录该内容

16 篇文章

订阅专栏

本文介绍了一种基于广度优先搜索的迷宫寻路算法，该算法能够在迷宫中找到从起点到终点的最短路径。文章详细解释了输入格式、约束条件及输出要求，并提供了一个实现该算法的C++程序示例。

【题目描述】

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

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’.

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

【数据样例】

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

2
6
-1
10

【我的程序】

#include <iostream>
#define arraySize 102
#define queueSize 10002
using namespace std;

int n,m;

int add_x[4]={0,1,0,-1};
int add_y[4]={1,0,-1,0};

int queueStart,queueEnd;
int queue_x[queueSize], queue_y[queueSize];

int visited[arraySize][arraySize];
int value[arraySize][arraySize];

char a[arraySize][arraySize];

int boundCheck(int x, int y)
{
    if (x<=n && x>0 && y<=m && y>0) return 1; else return 0;
}

void jouer()
{
    int x,y,x_neigh,y_neigh;
    char ch;

    cin >>n >>m;
    for (int i=1;i<=n;i++)
        for (int j=1;j<=m;j++)
    {
        cin >>ch;
        switch (ch)
        {
            case '.': value[i][j]=0; visited[i][j]=0; break;
            case '#': value[i][j]=-1; visited[i][j]=1; break;
            case 'S': queue_x[1]=i; queue_y[1]=j; value[i][j]=0; visited[i][j]=1; break;
            case 'E': value[i][j]=-1; visited[i][j]=0; break;
        }
        a[i][j]=ch;
    }

    queueStart=1; queueEnd=1;
    while (queueStart<=queueEnd)
    {
        x=queue_x[queueStart];
        y=queue_y[queueStart];
        queueStart++;

        for (int i=0;i<4;i++)
        {
            x_neigh=x+add_x[i];
            y_neigh=y+add_y[i];
            if (boundCheck(x_neigh,y_neigh) && visited[x_neigh][y_neigh]==0)
            {
                value[x_neigh][y_neigh]=value[x][y]+1;
                visited[x_neigh][y_neigh]=1;

                if (a[x_neigh][y_neigh]=='E') { cout <<value[x_neigh][y_neigh] <<endl; return;}

                queueEnd++;
                queue_x[queueEnd]=x_neigh; queue_y[queueEnd]=y_neigh;
            }
        }
    }

    cout <<-1 <<endl; return;
}

int main()
{
    int caseNum;
    cin >>caseNum;
    for (int i=0; i<caseNum; i++) jouer();
    return 0;
}