前往小程序,Get更优阅读体验!
立即前往
首页
学习
活动
专区
工具
TVP
发布
社区首页 >专栏 >2023-03-11:给定一个N*M的二维矩阵,只由字符'O'、'X'、'S'、'E'组成,'O'表示这个地方是可通行的平地,'

2023-03-11:给定一个N*M的二维矩阵,只由字符'O'、'X'、'S'、'E'组成,'O'表示这个地方是可通行的平地,'

作者头像
福大大架构师每日一题
发布2023-06-08 14:53:44
2370
发布2023-06-08 14:53:44
举报

2023-03-11:给定一个N*M的二维矩阵,只由字符'O'、'X'、'S'、'E'组成,

'O'表示这个地方是可通行的平地,

'X'表示这个地方是不可通行的障碍,

'S'表示这个地方有一个士兵,全图保证只有一个士兵,

'E'表示这个地方有一个敌人,全图保证只有一个敌人,

士兵可以在上、下、左、右四个方向上移动,

走到相邻的可通行的平地上,走一步耗费a个时间单位,

士兵从初始地点行动时,不管去哪个方向,都不用耗费转向的代价,

但是士兵在行动途中,如果需要转向,需要额外再付出b个时间单位。

返回士兵找到敌人的最少时间。

如果因为障碍怎么都找不到敌人,返回-1,

1 <= N,M <= 1000,

1 <= a,b <= 100000,

只会有一个士兵、一个敌人。

来自华为。

答案2023-03-11:

Dijkstra算法+优先级队列。

代码根据山寨版[chatgpt](https://chatgpt.zcorky.com/)稍做修改写的。这不得不承认chatgpt很强大,这还是山寨版的,感觉比我自己写得还要好。

以下代码是生成的rust代码,稍微做了修改。如下:

use rand::Rng;
use std::cmp::Reverse;
use std::collections::BinaryHeap;
fn random_map(n: usize, m: usize) -> Vec<Vec<char>> {
    let mut map = vec![vec!['O'; m]; n];
    for i in 0..n {
        for j in 0..m {
            if rand::thread_rng().gen_range(0, 2) == 0 {
                map[i][j] = 'X';
            }
        }
    }
    let si = rand::thread_rng().gen_range(0, n);
    let sj = rand::thread_rng().gen_range(0, m);
    map[si][sj] = 'S';
    let (mut ei, mut ej) = (si, sj);
    while ei == si && ej == sj {
        ei = rand::thread_rng().gen_range(0, n);
        ej = rand::thread_rng().gen_range(0, m);
    }
    map[ei][ej] = 'E';
    map
}

fn main() {
    let n = 3;
    let m = 4;
    let v = 10;
    println!("功能测试开始");
    for _ in 0..2000 {
        let map = random_map(n, m);
        let a = rand::thread_rng().gen_range(1, v + 1);
        let b = rand::thread_rng().gen_range(1, v + 1);
        let ans1 = min_cost1(&map, a, b);
        let ans2 = min_cost2(&map, a, b);
        if ans1 != ans2 {
            println!("出错了");
            println!("{}", ans1);
            println!("{}", ans2);
            return;
        }
    }
    println!("功能测试结束");

    println!("性能测试开始");
    let n = 1000;
    let m = 1000;
    let v = 100000;
    let a = rand::thread_rng().gen_range(1, v + 1);
    let b = rand::thread_rng().gen_range(1, v + 1);
    let map = random_map(n, m);
    println!("数据规模 : {} * {}", n, m);
    println!("通行代价 : {}", a);
    println!("转向代价 : {}", b);
    let start = std::time::Instant::now();
    min_cost2(&map, a, b);
    let end = std::time::Instant::now();
    println!("运行时间 : {}毫秒", (end - start).as_millis());
    println!("功能测试结束");
}

fn min_cost1(map: &[Vec<char>], a: i32, b: i32) -> i32 {
    let n = map.len();
    let m = map[0].len();
    let mut start_x = 0;
    let mut start_y = 0;
    for i in 0..n {
        for j in 0..m {
            if map[i][j] == 'S' {
                start_x = i;
                start_y = j;
            }
        }
    }
    let mut visited = vec![vec![vec![false; 4]; m]; n];
    let p1 = f(&map, start_x, start_y, 0, a, b, &mut visited);
    let p2 = f(&map, start_x, start_y, 1, a, b, &mut visited);
    let p3 = f(&map, start_x, start_y, 2, a, b, &mut visited);
    let p4 = f(&map, start_x, start_y, 3, a, b, &mut visited);
    let ans = p1.min(p2).min(p3).min(p4);
    if ans == i32::MAX {
        -1
    } else {
        ans - a
    }
}

fn f(
    map: &[Vec<char>],
    si: usize,
    sj: usize,
    d: usize,
    a: i32,
    b: i32,
    visited: &mut Vec<Vec<Vec<bool>>>,
) -> i32 {
    let n = map.len();
    let m = map[0].len();
    if si >= n || sj >= m || map[si][sj] == 'X' || visited[si][sj][d] {
        return i32::MAX;
    }
    if map[si][sj] == 'E' {
        return a;
    }
    visited[si][sj][d] = true;
    let p0 = f(&map, si.checked_sub(1).unwrap_or(0), sj, 0, a, b, visited);
    let p1 = f(&map, si + 1, sj, 1, a, b, visited);
    let p2 = f(&map, si, sj.checked_sub(1).unwrap_or(0), 2, a, b, visited);
    let p3 = f(&map, si, sj + 1, 3, a, b, visited);
    let p0 = if d != 0 && p0 != i32::MAX { p0 + b } else { p0 };
    let p1 = if d != 1 && p1 != i32::MAX { p1 + b } else { p1 };
    let p2 = if d != 2 && p2 != i32::MAX { p2 + b } else { p2 };
    let p3 = if d != 3 && p3 != i32::MAX { p3 + b } else { p3 };
    let ans = p0.min(p1).min(p2).min(p3);
    visited[si][sj][d] = false;
    if ans == i32::MAX {
        ans
    } else {
        ans + a
    }
}

fn min_cost2(map: &[Vec<char>], a: i32, b: i32) -> i32 {
    let n = map.len();
    let m = map[0].len();
    let mut start_x = 0;
    let mut start_y = 0;
    for i in 0..n {
        for j in 0..m {
            if map[i][j] == 'S' {
                start_x = i;
                start_y = j;
            }
        }
    }
    let mut heap = BinaryHeap::new();
    heap.push((Reverse(0), start_x, start_y, 0));
    heap.push((Reverse(0), start_x, start_y, 1));
    heap.push((Reverse(0), start_x, start_y, 2));
    heap.push((Reverse(0), start_x, start_y, 3));
    // (i,j,朝向)
    let mut visited = vec![vec![vec![false; 4]; m]; n];
    let mut ans = -1;
    while let Some((Reverse(cost), x, y, direction)) = heap.pop() {
        if visited[x][y][direction] {
            continue;
        }
        if map[x][y] == 'E' {
            ans = cost;
            break;
        }
        visited[x][y][direction] = true;
        add(
            x as i32 - 1,
            y as i32,
            0,
            direction,
            cost,
            a,
            b,
            map,
            &mut visited,
            &mut heap,
        );
        add(
            x as i32 + 1,
            y as i32,
            1,
            direction,
            cost,
            a,
            b,
            map,
            &mut visited,
            &mut heap,
        );
        add(
            x as i32,
            y as i32 - 1,
            2,
            direction,
            cost,
            a,
            b,
            map,
            &mut visited,
            &mut heap,
        );
        add(
            x as i32,
            y as i32 + 1,
            3,
            direction,
            cost,
            a,
            b,
            map,
            &mut visited,
            &mut heap,
        );
    }
    ans
}

// 从(x,y, pre_d) -> (i,j,d)
// 走格子的代价a
// 转向的代价是b
// pre_c + a
fn add(
    i: i32,
    j: i32,
    direction: usize,
    pre_direction: usize,
    pre_cost: i32,
    a: i32,
    b: i32,
    map: &[Vec<char>],
    visited: &mut Vec<Vec<Vec<bool>>>,
    heap: &mut BinaryHeap<(Reverse<i32>, usize, usize, usize)>,
) {
    let n = map.len() as i32;
    let m = map[0].len() as i32;
    if i < 0
        || i >= n
        || j < 0
        || j >= m
        || map[i as usize][j as usize] == 'X'
        || visited[i as usize][j as usize][direction]
    {
        return;
    }
    let mut cost = pre_cost + a;
    if direction != pre_direction {
        cost += b;
    }
    heap.push((Reverse(cost), i as usize, j as usize, direction));
}

以下代码是生成的golang代码,稍微做了修改。如下:

package main

import (
  "container/heap"
  "fmt"
  "math/rand"
  "time"
)

func minCost1(mapData [][]byte, a int, b int) int {
  // 获取地图大小和起点位置
  n, m := len(mapData), len(mapData[0])
  startX, startY := 0, 0
  for i := 0; i < n; i++ {
    for j := 0; j < m; j++ {
      if mapData[i][j] == 'S' {
        startX, startY = i, j
        break
      }
    }
  }

  // 初始化 visited 数组
  visited := make([][][]bool, n)
  for i := range visited {
    visited[i] = make([][]bool, m)
    for j := range visited[i] {
      visited[i][j] = make([]bool, 4)
    }
  }

  // 计算从四个方向到达终点的最短距离
  p1 := f(mapData, startX, startY, 0, a, b, visited)
  p2 := f(mapData, startX, startY, 1, a, b, visited)
  p3 := f(mapData, startX, startY, 2, a, b, visited)
  p4 := f(mapData, startX, startY, 3, a, b, visited)

  // 返回四个方向中最小的距离
  ans := min(p1, min(p2, min(p3, p4)))
  if ans == 1<<31-1 {
    return -1
  } else {
    return ans - a
  }
}

func f(mapData [][]byte, si int, sj int, d int, a int, b int, visited [][][]bool) int {
  // 如果出现越界、墙壁或者已经访问过的情况,返回一个大整数表示无法到达该位置
  if si < 0 || si == len(mapData) || sj < 0 || sj == len(mapData[0]) || mapData[si][sj] == 'X' || visited[si][sj][d] {
    return 1<<31 - 1
  }

  // 如果到达终点,返回 a 表示到达终点所需的代价
  if mapData[si][sj] == 'E' {
    return a
  }

  // 标记该位置已经被访问过
  visited[si][sj][d] = true

  // 计算从四个方向到达下一个位置所需的代价(如果可以到达的话)
  var p [4]int
  p[0] = f(mapData, si-1, sj, 0, a, b, visited)
  p[1] = f(mapData, si+1, sj, 1, a, b, visited)
  p[2] = f(mapData, si, sj-1, 2, a, b, visited)
  p[3] = f(mapData, si, sj+1, 3, a, b, visited)
  if d != 0 && p[0] != 1<<31-1 {
    p[0] += b
  }
  if d != 1 && p[1] != 1<<31-1 {
    p[1] += b
  }
  if d != 2 && p[2] != 1<<31-1 {
    p[2] += b
  }
  if d != 3 && p[3] != 1<<31-1 {
    p[3] += b
  }

  // 返回四个方向中最小的代价,并且取消对该位置的访问标记
  ans := min(p[0], min(p[1], min(p[2], p[3])))
  visited[si][sj][d] = false
  if ans == 1<<31-1 {
    return ans
  } else {
    return ans + a
  }
}

func min(a int, b int) int {
  if a < b {
    return a
  } else {
    return b
  }
}

func minCost2(mapData [][]byte, a int, b int) int {
  // 获取地图大小和起点位置
  n, m := len(mapData), len(mapData[0])
  startX, startY := 0, 0
  for i := 0; i < n; i++ {
    for j := 0; j < m; j++ {
      if mapData[i][j] == 'S' {
        startX, startY = i, j
        break
      }
    }
  }

  // 初始化优先队列和 visited 数组
  h := &minHeap{}
  heap.Init(h)
  heap.Push(h, node{startX, startY, 0, 0})
  heap.Push(h, node{startX, startY, 1, 0})
  heap.Push(h, node{startX, startY, 2, 0})
  heap.Push(h, node{startX, startY, 3, 0})

  visited := make([][][]bool, n)
  for i := range visited {
    visited[i] = make([][]bool, m)
    for j := range visited[i] {
      visited[i][j] = make([]bool, 4)
    }
  }

  // 运行 Dijkstra 算法并返回答案
  ans := -1
  for h.Len() > 0 {
    cur := heap.Pop(h).(node)
    if visited[cur.x][cur.y][cur.dir] {
      continue
    }
    if mapData[cur.x][cur.y] == 'E' {
      ans = cur.cost
      break
    }
    visited[cur.x][cur.y][cur.dir] = true
    add(cur.x-1, cur.y, 0, cur.dir, cur.cost, a, b, mapData, visited, h)
    add(cur.x+1, cur.y, 1, cur.dir, cur.cost, a, b, mapData, visited, h)
    add(cur.x, cur.y-1, 2, cur.dir, cur.cost, a, b, mapData, visited, h)
    add(cur.x, cur.y+1, 3, cur.dir, cur.cost, a, b, mapData, visited, h)
  }

  return ans
}

type node struct {
  x, y, dir, cost int
}

type minHeap []node

func (h minHeap) Len() int { return len(h) }

func (h minHeap) Less(i, j int) bool { return h[i].cost < h[j].cost }

func (h minHeap) Swap(i, j int) { h[i], h[j] = h[j], h[i] }

func (h *minHeap) Push(x interface{}) { *h = append(*h, x.(node)) }

func (h *minHeap) Pop() interface{} {
  old := *h
  n := len(old)
  x := old[n-1]
  *h = old[:n-1]
  return x
}

func add(i, j, d, preD, preC, a, b int, mapData [][]byte, visited [][][]bool, h *minHeap) {
  if i < 0 || i == len(mapData) || j < 0 || j == len(mapData[0]) || mapData[i][j] == 'X' || visited[i][j][d] {
    return
  }
  cost := preC + a
  if d != preD {
    cost += b
  }
  heap.Push(h, node{i, j, d, cost})
}

func randomMap(n, m int) [][]byte {
  mapData := make([][]byte, n)
  for i := range mapData {
    mapData[i] = make([]byte, m)
    for j := range mapData[i] {
      if rand.Float32() < 0.5 {
        mapData[i][j] = 'O'
      } else {
        mapData[i][j] = 'X'
      }
    }
  }
  si := rand.Intn(n)
  sj := rand.Intn(m)
  mapData[si][sj] = 'S'
  var ei, ej int
  for {
    ei = rand.Intn(n)
    ej = rand.Intn(m)
    if ei != si || ej != sj {
      break
    }
  }
  mapData[ei][ej] = 'E'
  return mapData
}

func main() {
  n, m := 3, 4
  v := 10
  fmt.Println("功能测试开始")
  for i := 0; i < 2000; i++ {
    mapData := randomMap(n, m)
    a := rand.Intn(v) + 1
    b := rand.Intn(v) + 1
    ans1 := minCost1(mapData, a, b)
    ans2 := minCost2(mapData, a, b)
    if ans1 != ans2 {
      fmt.Println("出错了")
      fmt.Println(ans1)
      fmt.Println(ans2)
    }
  }
  fmt.Println("功能测试结束")

  fmt.Println("性能测试开始")
  n = 1000
  m = 1000
  v = 100000
  a := rand.Intn(v) + 1
  b := rand.Intn(v) + 1
  mapData := randomMap(n, m)
  fmt.Println("数据规模 : ", n, " * ", m)
  fmt.Println("通行代价 : ", a)
  fmt.Println("转向代价 : ", b)
  start := time.Now()
  minCost2(mapData, a, b)
  end := time.Now()
  fmt.Println("运行时间 : ", end.Sub(start))
  fmt.Println("功能测试结束")
}
本文参与 腾讯云自媒体分享计划,分享自微信公众号。
原始发表:2023-03-11,如有侵权请联系 cloudcommunity@tencent.com 删除

本文分享自 福大大架构师每日一题 微信公众号,前往查看

如有侵权,请联系 cloudcommunity@tencent.com 删除。

本文参与 腾讯云自媒体分享计划  ,欢迎热爱写作的你一起参与!

评论
登录后参与评论
0 条评论
热度
最新
推荐阅读
领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档