Golang算法模板-bfs模板+路径记忆

2019蓝桥杯大学B组E题

题目描述

<https://blog.csdn.net/weixin_42172261/article/details/89647625

https://blog.csdn.net/a1097304791/article/details/88791546

package main

import (
	"fmt"
)

var next [][]int = [][]int{{1, 0}, {0, -1}, {0, 1}, {-1, 0}} //字典序bfs
var path []byte = []byte{'D', 'L', 'R', 'U'}
var maze [100][100]int //迷宫
var book [100][100]int //标记当前是否走过
var n, m int

type node struct {
	x, y, step, pre int //pre代表上一步在队列q中的下标
	ch              byte
}

var q [10000]node //bfs队列，出队不销毁
func getpath(t int) {
	if t == 1 {
		return
	}
	getpath(q[t].pre)
	fmt.Printf("%c", q[t].ch)
}
func bfs() {
	head, tail := 1, 1
	q[tail].x, q[tail].y = 0, 0
	q[tail].step, q[tail].pre = 0, -1
	tail++
	book[0][0] = 1

	flag := false
	for head < tail {
		for k := 0; k < 4; k++ {
			tx := q[head].x + next[k][0]
			ty := q[head].y + next[k][1]
			if tx < 0 || tx >= n || ty < 0 || ty >= m || book[tx][ty] != 0 || maze[tx][ty] == 1 {
				continue
			}
			book[tx][ty] = 1
			q[tail].x, q[tail].y = tx, ty
			q[tail].pre = head
			q[tail].step = q[head].step + 1
			q[tail].ch = path[k]
			tail++
			if tx == n-1 && ty == m-1 {
				flag = true
				break
			}
		}
		if flag == true {
			break
		}
		head++
	}
	getpath(tail - 1)
}
func main() {
	fmt.Scanf("%d %d\n", &n, &m)
	for i := 0; i < n; i++ {
		var s1 string
		fmt.Scanf("%s\n", &s1)
		//fmt.Printf(":%s\n", s1)
		//fmt.Printf("%d, %s\n", len(s1), s1)
		for j := 0; j < m; j++ {
			maze[i][j] = int(s1[j] - '0')
		}
	}
	bfs()
}