Golang颜色轮问题解决方案探讨

这是一个经典的图着色问题变体，可以用图论和约束满足算法解决。以下是Go语言的实现方案：

package main

import (
	"fmt"
	"sort"
)

type Region struct {
	id       int
	color    string
	adjacent []int
}

func solveColorWheel(regions []Region, colors []string) (map[int]string, bool) {
	// 按相邻区域数量排序，优先处理约束多的区域
	sort.Slice(regions, func(i, j int) bool {
		return len(regions[i].adjacent) > len(regions[j].adjacent)
	})
	
	colorMap := make(map[int]string)
	regionMap := make(map[int]Region)
	for _, r := range regions {
		regionMap[r.id] = r
	}
	
	return backtrack(regions, colors, colorMap, regionMap, 0)
}

func backtrack(regions []Region, colors []string, colorMap map[int]string, 
	regionMap map[int]Region, index int) (map[int]string, bool) {
	if index == len(regions) {
		return colorMap, true
	}
	
	current := regions[index]
	
	// 获取可用颜色
	available := getAvailableColors(current, colorMap, regionMap, colors)
	
	for _, color := range available {
		colorMap[current.id] = color
		
		if solution, ok := backtrack(regions, colors, colorMap, regionMap, index+1); ok {
			return solution, true
		}
		
		delete(colorMap, current.id)
	}
	
	return nil, false
}

func getAvailableColors(region Region, colorMap map[int]string, 
	regionMap map[int]Region, colors []string) []string {
	used := make(map[string]bool)
	
	// 检查相邻区域已使用的颜色
	for _, adjID := range region.adjacent {
		if adjColor, exists := colorMap[adjID]; exists {
			used[adjColor] = true
		}
	}
	
	// 收集可用颜色
	var available []string
	for _, color := range colors {
		if !used[color] {
			available = append(available, color)
		}
	}
	
	return available
}

func main() {
	// 示例：6个区域的色轮问题
	regions := []Region{
		{id: 1, adjacent: []int{2, 6}},
		{id: 2, adjacent: []int{1, 3}},
		{id: 3, adjacent: []int{2, 4}},
		{id: 4, adjacent: []int{3, 5}},
		{id: 5, adjacent: []int{4, 6}},
		{id: 6, adjacent: []int{5, 1}},
	}
	
	colors := []string{"红", "绿", "蓝"}
	
	solution, found := solveColorWheel(regions, colors)
	if found {
		fmt.Println("找到解决方案：")
		for id, color := range solution {
			fmt.Printf("区域 %d -> %s\n", id, color)
		}
	} else {
		fmt.Println("无解")
	}
}

对于更复杂的色轮问题（如链接中的12区域问题），可以使用带剪枝的回溯算法：

func solveComplexWheel() {
	regions := []Region{
		{id: 1, adjacent: []int{2, 12}},
		{id: 2, adjacent: []int{1, 3, 9}},
		{id: 3, adjacent: []int{2, 4}},
		{id: 4, adjacent: []int{3, 5, 11}},
		{id: 5, adjacent: []int{4, 6}},
		{id: 6, adjacent: []int{5, 7, 10}},
		{id: 7, adjacent: []int{6, 8}},
		{id: 8, adjacent: []int{7, 9, 12}},
		{id: 9, adjacent: []int{2, 8, 10}},
		{id: 10, adjacent: []int{6, 9, 11}},
		{id: 11, adjacent: []int{4, 10, 12}},
		{id: 12, adjacent: []int{1, 8, 11}},
	}
	
	colors := []string{"A", "B", "C", "D"}
	
	solution, found := solveColorWheel(regions, colors)
	if found {
		// 按区域ID排序输出
		var ids []int
		for id := range solution {
			ids = append(ids, id)
		}
		sort.Ints(ids)
		
		fmt.Println("12区域色轮解决方案：")
		for _, id := range ids {
			fmt.Printf("R%d: %s\n", id, solution[id])
		}
	}
}

这个实现使用了回溯算法，通过MRV（最少剩余值）启发式优化，优先处理约束最多的区域，提高了搜索效率。算法时间复杂度为O(c^n)，其中c是颜色数量，n是区域数量，但通过剪枝在实际问题中表现良好。