NYOJ 58 最少步数问题

迷宫最短路径算法

最新推荐文章于 2019-03-11 11:57:48 发布

原创最新推荐文章于 2019-03-11 11:57:48 发布 · 329 阅读

0 ·

CC 4.0 BY-SA版权

搜索同时被 2 个专栏收录

5 篇文章

订阅专栏

DFS

4 篇文章

订阅专栏

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)遍历迷宫地图，寻找从起点到终点的最短步数。文章提供了两种不同的C语言实现方式，并详细展示了如何使用递归进行路径探索。

//修改版

#include<stdio.h>
#include<string.h>
int arr[9][9] = {
  1,1,1,1,1,1,1,1,1,
  1,0,0,1,0,0,1,0,1,
  1,0,0,1,1,0,0,0,1,
  1,0,1,0,1,1,0,1,1,
  1,0,0,0,0,1,0,0,1,
  1,1,0,1,0,1,0,0,1,
  1,1,0,1,0,1,0,0,1,
  1,1,0,1,0,0,0,0,1,
  1,1,1,1,1,1,1,1,1,
};
int book[9][9];
int Next[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
int starx, stary, endx, endy;
int min_step = 999999;
void DFS(int x, int y,int step)
{
	if(x == endx && y == endy)
	{
		if(step < min_step)
		min_step = step;
		return ;
	}
	int i,tempx,tempy;
	for(i = 0;i<4;i++)
	{
		tempx = x + Next[i][0];
		tempy = y + Next[i][1];
		if(tempx < 1 || tempx > 8 || tempy < 1 || tempy > 8)
		continue;
		if(arr[tempx][tempy] == 0 && book[tempx][tempy] == 0)
		{
			book[tempx][tempy] = 1;
			DFS(tempx,tempy,step + 1);
			book[tempx][tempy] = 0;
		}
	}
	return ;
	
}
int main()
{
	int N;
	scanf("%d",&N);
	while(N--)
	{
		min_step = 999999;
		scanf("%d %d %d %d",&starx,&stary,&endx,&endy);
		memset(book, 0, sizeof(book));
		book[starx][stary] = 1;
		DFS(starx,stary,0);
		printf("%d\n",min_step);
	}
	return 0;
 }

//两个月后重新看这个感觉简单了很多，优化后：

#include<cstdio>
#define min(x,y) x>y?y:x
int arr[9][9] = {  
  1,1,1,1,1,1,1,1,1,  
  1,0,0,1,0,0,1,0,1,  
  1,0,0,1,1,0,0,0,1,  
  1,0,1,0,1,1,0,1,1,  
  1,0,0,0,0,1,0,0,1,  
  1,1,0,1,0,1,0,0,1,  
  1,1,0,1,0,1,0,0,1,  
  1,1,0,1,0,0,0,0,1,  
  1,1,1,1,1,1,1,1,1,  
};
int a, b, ea, eb, min_step;
void dfs(int x, int y, int step)
{
	if(arr[x][y])
	return ;
	if(x == ea && y == eb)
	{
		min_step = min(step,min_step);
		return ;
	}
	arr[x][y] = 1;
	dfs(x+1,y,step+1);
	dfs(x,y+1,step+1);
	dfs(x-1,y,step+1);
	dfs(x,y-1,step+1);
	arr[x][y] = 0;
}
int main()
{
	int n;
	scanf("%d",&n);
	while(n--)
	{
		min_step = 0xfffff;
		scanf("%d%d%d%d",&a,&b,&ea,&eb);
		dfs(a, b, 0);
		printf("%d\n",min_step);
	}
}