关于图的搜索问题可以看我的另一篇博客二分图的判定点击打开链接

L2-023. 图着色问题

6 8 3
2 1
1 3
4 6
2 5
2 4
5 4
5 6
3 6
4
1 2 3 3 1 2
4 5 6 6 4 5
1 2 3 4 5 6
2 3 4 2 3 4

Yes
Yes
No
No

我自己的解答最后一个测试点会超时，最后是我参考的别人的代码，觉得很巧妙

先来看一下最后一个点超时的代码，作参考

import java.io.BufferedInputStream;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Scanner;
import java.util.Set;

public class Main {
    public static int V, E, C;
    public static List<Integer>[] list;
    public static int[] color = new int[501];
    public static int[][] book; // 标记数组，避免往回走，成往返回路死在递归了

    public static int[] checked;  // 做剪枝用，如果某个顶点在之前相邻顶点的邻接数组用过，说明走过一遍了，就没必要走了

    public static void main(String[] args) {
        Scanner cin = new Scanner(new BufferedInputStream(System.in));
        V = cin.nextInt();
        E = cin.nextInt();
        C = cin.nextInt();
        list = (ArrayList<Integer>[]) new ArrayList[V + 1];
        for (int i = 1; i <= V; ++i) {
            list[i] = new ArrayList<Integer>();
        }
        book = new int[V + 1][V + 1];
        checked = new int[V + 1];
        for (int i = 0; i < E; ++i) {
            int a = cin.nextInt();
            int b = cin.nextInt();
            list[a].add(b);
            list[b].add(a);
        }
        int n = cin.nextInt();
        Set<Integer> set = new HashSet<Integer>();
        for (int i = 0; i < n; ++i) {
            set.clear();
            for (int j = 1; j <= V; ++j) {
                color[j] = cin.nextInt();
                set.add(color[j]);
            }
            if (set.size() != C) { // 用例的第三个点，3种颜色却给出6种，不合规范
                System.out.println("No");
                continue;
            }
            solve();
        }
        cin.close();
    }

    public static void solve() {
        clear();
        for (int i = 1; i <= V; ++i) {
            if (checked[i] != 1 && !dfs(i)) { // 题目没说图是连通的，所以要每个顶点走一遍
                System.out.println("No");
                return;
            }
        }
        System.out.println("Yes");
    }

    public static void clear() {
        for (int i = 0; i < V + 1; ++i) {
            for (int j = 0; j < V + 1; ++j) {
                book[i][j] = 0;
            }
        }
        for (int i = 0; i < V + 1; ++i) {
            checked[i] = 0;
        }
    }

    public static boolean dfs(int v) {
        int size = list[v].size();
        for (int i = 0; i < size; ++i) {
            int vv = list[v].get(i);
            checked[vv] = 1; // 下一次返回solve()可以剪枝
            if (book[v][vv] != 1) {
                if (color[vv] == color[v])
                    return false;
                book[v][vv] = 1;
                book[vv][v] = 1;
                if (!dfs(vv))
                    return false;
            }
        }
        return true;
    }
}

接着是正确的代码：

#include <bits/stdc++.h>

using namespace std;
int V, E, C;
int v[250000]; // 起点的顶点,每个点的终点可以是另外499个点
int w[250000]; // 终点的顶点
int color[501];
int visit[501]; // 辅助标记数组
void clear()
{
    for (int i = 0; i < 501; ++i)
	{
        visit[i] = 0;
    }
}
int main()
{
	scanf("%d %d %d", &V, &E, &C);
	for (int i = 1; i <= E; ++i)
	{
        scanf("%d %d", &v[i], &w[i]);
    }
    int n;
    scanf("%d", &n);
	for (int i = 0; i < n; ++i)
	{
	    int count = 0;
	    bool flag = true;
	    clear();
	    for (int j = 1; j <= V; ++j)
		{
	        scanf("%d", &color[j]);
	        if (visit[color[j]])
	            continue;
	        visit[color[j]] = 1; // 这个颜色已经有了
	        ++count; // 记录颜色种数
	    }
	    if (count != C) {
	        printf("No\n");
	        continue;
	    }
	
	    for (int k = 1; flag && k <= E; ++k) {
	        if (color[v[k]] == color[w[k]]) {
	            printf("No\n");
	            flag = false;
	        }
	    }
	    if (flag) {
	        printf("Yes\n");
	    }
	}
	return 0;
}

题目链接地址https://www.patest.cn/contests/gplt/L2-023

========================================Talk is cheap, show me the code=======================================