2024-04-03:用go语言,在一个小城市里,有 m 个房子排成一排, 你需要给每个房子涂上 n 种颜色之一

2024-04-03:用go语言,在一个小城市里,有 m 个房子排成一排,

你需要给每个房子涂上 n 种颜色之一(颜色编号为 1 到 n ),

有的房子去年夏天已经涂过颜色了,所以这些房子不可以被重新涂色,

我们将连续相同颜色尽可能多的房子称为一个街区。

比方说 houses = [1,2,2,3,3,2,1,1],

它包含 5 个街区 [{1}, {2,2}, {3,3}, {2}, {1,1}]。

给你一个数组 houses ,一个 m * n 的矩阵 cost 和一个整数 target,其中:

houses[i]:是第 i 个房子的颜色,0 表示这个房子还没有被涂色,

cost[i][j]:是将第 i 个房子涂成颜色 j+1 的花费。

请你返回房子涂色方案的最小总花费,使得每个房子都被涂色后,恰好组成 target 个街区。

如果没有可用的涂色方案,请返回 -1。

输入:houses = [0,0,0,0,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3。

输出:9。

答案2024-04-03:

来自左程云

灵捷3.5

大体步骤如下:

minCost1函数:

1.首先,创建一个三维数组dp,用于记录状态转移的结果。dp[i][k][c]表示将前i个房子涂色,形成k个街区,并且第i个房子颜色为c+1时的最小总花费。

2.然后,调用process1函数,进行递归计算。

3.在process1函数中,首先处理边界情况:如果k小于0,返回无穷大;如果i等于房子数量,如果k等于0,返回0,否则返回无穷大。

4.如果dp[i][k][c]已经计算过,直接返回结果。

5.接着,判断第i个房子的颜色是否已经确定(houses[i] != 0):

  • 如果颜色已经确定,判断该颜色是否与c相同,分别递归处理下一个房子(i+1)和剩余街区数量(k-1或k)。

  • 如果颜色未确定,遍历每种可能的颜色(1到n),分别递归处理下一个房子和剩余街区数量,并记录选择花费最小的方案。

6.将结果存入dp[i][k][c],并返回结果。

minCost2函数:

1.创建一个二维数组dp,用于记录状态转移的结果。dp[k][c]表示形成k个街区,且最后一个房子颜色为c+1时的最小总花费。

2.首先初始化dp数组:对于k=0(没有街区)和所有的颜色c,花费为0;对于k>0和所有的颜色c,花费初始化为无穷大。

3.然后,从后向前遍历房子,处理每个房子的情况:

  • 如果房子颜色已经确定(houses[i] != 0),更新dp数组中对应位置的值。

  • 如果房子颜色未确定,通过dp数组中记录的上一次的结果,计算每个街区数量和颜色的最小花费,更新dp数组中对应位置的值。

4.最后,返回dp[target][0]的结果,如果为无穷大则返回-1。

minCost3函数:

1.创建一个二维数组dp,用于记录状态转移的结果。dp[k][c]表示形成k个街区,且最后一个房子颜色为c+1时的最小总花费。

2.首先初始化dp数组和辅助数组minl和minr:

  • 对于k=0(没有街区)和所有的颜色c,花费为0;

  • 对于k>0和所有的颜色c,花费初始化为无穷大;

  • minl[i]表示前i个颜色中最小的花费,minr[i]表示从第i个颜色到第n个颜色中最小的花费。

3.然后,从后向前遍历房子,处理每个房子的情况:

  • 如果房子颜色已经确定(houses[i] != 0),更新dp数组中对应位置的值。

  • 如果房子颜色未确定,通过dp数组中记录的上一次的结果和辅助数组minl和minr,计算每个街区数量和颜色的最小花费,更新dp数组中对应位置的值。

4.最后,返回dp[target][0]的结果,如果为无穷大则返回-1。

这3种算法的时间复杂度和空间复杂度如下:

  • minCost1:时间复杂度为O(m * n^target),空间复杂度为O(m * target * n)。

  • minCost2:时间复杂度为O(m * target * n^2),空间复杂度为O(target * n)。

  • minCost3:时间复杂度为O(m * target * n^2),空间复杂度为O(n)。

Go完整代码如下:

package main

import (
    "fmt"
    "math"
)

func minCost1(houses []int, cost [][]int, m int, n int, target int) int {
    dp := make([][][]int, m)
    for i := 0; i < m; i++ {
        dp[i] = make([][]int, target+1)
        for k := 0; k <= target; k++ {
            dp[i][k] = make([]int, n+1)
            for c := 0; c <= n; c++ {
                dp[i][k][c] = -1
            }
        }
    }
    ans := process1(houses, cost, n, 0, target, 0, dp)
    if ans == math.MaxInt32 {
        return -1
    }
    return ans
}

func process1(houses []int, cost [][]int, n, i, k, c int, dp [][][]int) int {
    if k < 0 {
        return math.MaxInt32
    }
    if i == len(houses) {
        if k == 0 {
            return 0
        }
        return math.MaxInt32
    }
    if dp[i][k][c] != -1 {
        return dp[i][k][c]
    }
    ans := math.MaxInt32
    if houses[i] != 0 {
        if houses[i] != c {
            ans = process1(houses, cost, n, i+1, k-1, houses[i], dp)
        } else {
            ans = process1(houses, cost, n, i+1, k, houses[i], dp)
        }
    } else {
        for fill := 1; fill <= n; fill++ {
            var next int
            if fill == c {
                next = process1(houses, cost, n, i+1, k, fill, dp)
            } else {
                next = process1(houses, cost, n, i+1, k-1, fill, dp)
            }
            if next != math.MaxInt32 {
                ans = min(ans, cost[i][fill-1]+next)
            }
        }
    }
    dp[i][k][c] = ans
    return ans
}

func minCost2(houses []int, cost [][]int, m int, n int, target int) int {
    dp := make([][]int, target+1)
    for k := range dp {
        dp[k] = make([]int, n+1)
    }
    for c := 0; c <= n; c++ {
        dp[0][c] = 0
    }
    for k := 1; k <= target; k++ {
        for c := 0; c <= n; c++ {
            dp[k][c] = math.MaxInt32
        }
    }
    memo := make([]int, n+1)
    for i := m - 1; i >= 0; i-- {
        if houses[i] != 0 {
            houseColor := houses[i]
            for k := target; k >= 0; k-- {
                memory := dp[k][houseColor]
                for c := 0; c <= n; c++ {
                    if houseColor != c {
                        if k == 0 {
                            dp[k][c] = math.MaxInt32
                        } else {
                            dp[k][c] = dp[k-1][houseColor]
                        }
                    } else {
                        dp[k][c] = memory
                    }
                }
            }
        } else {
            for k := target; k >= 0; k-- {
                for c := 0; c <= n; c++ {
                    memo[c] = dp[k][c]
                }
                for c := 0; c <= n; c++ {
                    ans := math.MaxInt32
                    for fill := 1; fill <= n; fill++ {
                        var next int
                        if fill == c {
                            next = memo[fill]
                        } else {
                            if k == 0 {
                                next = math.MaxInt32
                            } else {
                                next = dp[k-1][fill]
                            }
                        }
                        if next != math.MaxInt32 {
                            ans = min(ans, cost[i][fill-1]+next)
                        }
                    }
                    dp[k][c] = ans
                }
            }
        }
    }
    if dp[target][0] == math.MaxInt32 {
        return -1
    }
    return dp[target][0]
}

func minCost3(houses []int, cost [][]int, m int, n int, target int) int {
    dp := make([][]int, target+1)
    for k := range dp {
        dp[k] = make([]int, n+1)
    }
    for c := 0; c <= n; c++ {
        dp[0][c] = 0
    }
    for k := 1; k <= target; k++ {
        for c := 0; c <= n; c++ {
            dp[k][c] = math.MaxInt32
        }
    }
    memo := make([]int, n+1)
    minl := make([]int, n+2)
    minr := make([]int, n+2)
    minr[n+1] = math.MaxInt32
    minl[n+1] = minr[n+1]
    minr[0] = minl[n+1]
    minl[0] = minr[0]
    for i := m - 1; i >= 0; i-- {
        if houses[i] != 0 {
            for k, memory := target, 0; k >= 0; k-- {
                memory = dp[k][houses[i]]
                for c := 0; c <= n; c++ {
                    if houses[i] != c {
                        if k == 0 {
                            dp[k][c] = math.MaxInt32
                        } else {
                            dp[k][c] = dp[k-1][houses[i]]
                        }
                    } else {
                        dp[k][c] = memory
                    }
                }
            }
        } else {
            for k := target; k >= 0; k-- {
                for c := 0; c <= n; c++ {
                    memo[c] = dp[k][c]
                }
                for fill := 1; fill <= n; fill++ {
                    if k == 0 || dp[k-1][fill] == math.MaxInt32 {
                        minl[fill] = minl[fill-1]
                    } else {
                        minl[fill] = min(minl[fill-1], cost[i][fill-1]+dp[k-1][fill])
                    }
                }
                for fill := n; fill >= 1; fill-- {
                    if k == 0 || dp[k-1][fill] == math.MaxInt32 {
                        minr[fill] = minr[fill+1]
                    } else {
                        minr[fill] = min(minr[fill+1], cost[i][fill-1]+dp[k-1][fill])
                    }
                }
                for c, ans := 0, 0; c <= n; c++ {
                    if c == 0 || memo[c] == math.MaxInt32 {
                        ans = math.MaxInt32
                    } else {
                        ans = cost[i][c-1] + memo[c]
                    }
                    if c > 0 {
                        ans = min(ans, minl[c-1])
                    }
                    if c < n {
                        ans = min(ans, minr[c+1])
                    }
                    dp[k][c] = ans
                }
            }
        }
    }
    if dp[target][0] != math.MaxInt32 {
        return dp[target][0]
    }
    return -1
}

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

func main() {
    houses := []int{0, 0, 0, 0, 0}
    cost := [][]int{{1, 10}, {10, 1}, {10, 1}, {1, 10}, {5, 1}}
    m := 5
    n := 2
    target := 3

    fmt.Println(minCost1(houses, cost, m, n, target))
    fmt.Println(minCost2(houses, cost, m, n, target))
    fmt.Println(minCost3(houses, cost, m, n, target))
}

在这里插入图片描述

Python完整代码如下:

# -*-coding:utf-8-*-

import sys

def minCost1(houses, cost, m, n, target):
    dp = [[[-1] * (n + 1) for _ in range(target + 1)] for _ in range(m)]
    ans = process1(houses, cost, n, 0, target, 0, dp)
    if ans == sys.maxsize:
        return -1
    return ans

def process1(houses, cost, n, i, k, c, dp):
    if k < 0:
        return sys.maxsize
    if i == len(houses):
        if k == 0:
            return 0
        return sys.maxsize
    if dp[i][k][c] != -1:
        return dp[i][k][c]
    ans = sys.maxsize
    if houses[i] != 0:
        if houses[i] != c:
            ans = process1(houses, cost, n, i + 1, k - 1, houses[i], dp)
        else:
            ans = process1(houses, cost, n, i + 1, k, houses[i], dp)
    else:
        for fill in range(1, n + 1):
            if fill == c:
                next = process1(houses, cost, n, i + 1, k, fill, dp)
            else:
                next = process1(houses, cost, n, i + 1, k - 1, fill, dp)
            if next != sys.maxsize:
                ans = min(ans, cost[i][fill - 1] + next)
    dp[i][k][c] = ans
    return ans

def minCost2(houses, cost, m, n, target):
    dp = [[0] * (n + 1) for _ in range(target + 1)]
    for c in range(n + 1):
        dp[0][c] = 0
    for k in range(1, target + 1):
        for c in range(n + 1):
            dp[k][c] = sys.maxsize
    memo = [0] * (n + 1)
    for i in range(m - 1, -1, -1):
        if houses[i] != 0:
            houseColor = houses[i]
            for k in range(target, -1, -1):
                memory = dp[k][houseColor]
                for c in range(n + 1):
                    if houseColor != c:
                        if k == 0:
                            dp[k][c] = sys.maxsize
                        else:
                            dp[k][c] = dp[k - 1][houseColor]
                    else:
                        dp[k][c] = memory
        else:
            for k in range(target, -1, -1):
                for c in range(n + 1):
                    memo[c] = dp[k][c]
                for c in range(n + 1):
                    ans = sys.maxsize
                    for fill in range(1, n + 1):
                        if fill == c:
                            next = memo[fill]
                        else:
                            if k == 0:
                                next = sys.maxsize
                            else:
                                next = dp[k - 1][fill]
                        if next != sys.maxsize:
                            ans = min(ans, cost[i][fill - 1] + next)
                    dp[k][c] = ans
    if dp[target][0] == sys.maxsize:
        return -1
    return dp[target][0]

def minCost3(houses, cost, m, n, target):
    dp = [[0] * (n + 1) for _ in range(target + 1)]
    for c in range(n + 1):
        dp[0][c] = 0
    for k in range(1, target + 1):
        for c in range(n + 1):
            dp[k][c] = sys.maxsize
    memo = [0] * (n + 1)
    minl = [sys.maxsize] * (n + 2)
    minr = [sys.maxsize] * (n + 2)
    minl[n + 1] = minr[n + 1] = sys.maxsize
    minl[0] = minr[0] = sys.maxsize
    for i in range(m - 1, -1, -1):
        if houses[i] != 0:
            for k in range(target, -1, -1):
                memory = dp[k][houses[i]]
                for c in range(n + 1):
                    if houses[i] != c:
                        if k == 0:
                            dp[k][c] = sys.maxsize
                        else:
                            dp[k][c] = dp[k - 1][houses[i]]
                    else:
                        dp[k][c] = memory
        else:
            for k in range(target, -1, -1):
                for c in range(n + 1):
                    memo[c] = dp[k][c]
                for fill in range(1, n + 1):
                    if k == 0 or dp[k - 1][fill] == sys.maxsize:
                        minl[fill] = minl[fill - 1]
                    else:
                        minl[fill] = min(minl[fill - 1], cost[i][fill - 1] + dp[k - 1][fill])
                for fill in range(n, 0, -1):
                    if k == 0 or dp[k - 1][fill] == sys.maxsize:
                        minr[fill] = minr[fill + 1]
                    else:
                        minr[fill] = min(minr[fill + 1], cost[i][fill - 1] + dp[k - 1][fill])
                for c in range(n + 1):
                    if c == 0 or memo[c] == sys.maxsize:
                        ans = sys.maxsize
                    else:
                        ans = cost[i][c - 1] + memo[c]
                    if c > 0:
                        ans = min(ans, minl[c - 1])
                    if c < n:
                        ans = min(ans, minr[c + 1])
                    dp[k][c] = ans
    if dp[target][0] != sys.maxsize:
        return dp[target][0]
    return -1

def min(a, b):
    return a if a < b else b

houses = [0, 0, 0, 0, 0]
cost = [[1, 10], [10, 1], [10, 1], [1, 10], [5, 1]]
m = 5
n = 2
target = 3

print(minCost1(houses, cost, m, n, target))
print(minCost2(houses, cost, m, n, target))
print(minCost3(houses, cost, m, n, target))

在这里插入图片描述

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