• 【PAT甲级 - C++题解】1154 Vertex Coloring


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    📚专栏地址:PAT题解集合
    📝原题地址:题目详情 - 1154 Vertex Coloring (pintia.cn)
    🔑中文翻译:顶点着色
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    1154 Vertex Coloring

    A proper vertex coloring is a labeling of the graph’s vertices with colors such that no two vertices sharing the same edge have the same color. A coloring using at most k colors is called a (proper) k-coloring.

    Now you are supposed to tell if a given coloring is a proper k-coloring.

    Input Specification:

    Each input file contains one test case. For each case, the first line gives two positive integers N and M (both no more than 104), being the total numbers of vertices and edges, respectively. Then M lines follow, each describes an edge by giving the indices (from 0 to N−1) of the two ends of the edge.

    After the graph, a positive integer K (≤ 100) is given, which is the number of colorings you are supposed to check. Then K lines follow, each contains N colors which are represented by non-negative integers in the range of int. The i-th color is the color of the i-th vertex.

    Output Specification:

    For each coloring, print in a line k-coloring if it is a proper k-coloring for some positive k, or No if not.

    Sample Input:

    10 11
    8 7
    6 8
    4 5
    8 4
    8 1
    1 2
    1 4
    9 8
    9 1
    1 0
    2 4
    4
    0 1 0 1 4 1 0 1 3 0
    0 1 0 1 4 1 0 1 0 0
    8 1 0 1 4 1 0 5 3 0
    1 2 3 4 5 6 7 8 8 9
    
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    Sample Output:

    4-coloring
    No
    6-coloring
    No
    
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    题意

    一个合适的顶点着色是指用各种颜色标记图中各个顶点,使得每条边的两个端点的颜色都不相同。

    如果一种合适的顶点着色方案使用了一共 k 种不同的颜色,则称其为合适的 k 着色(k-coloring)。

    现在,你需要判断给定的着色方案是否是合适的 k 着色方案。

    思路
    1. 直接用一个结构体数组 e 来存储每一条边。
    2. 输入每种染色方案后,遍历每一条边,判断是否每一条边都满足两个点的颜色都不相同,只要有一个边出现两点颜色相同,则都不是一个合法的着色方案。
    3. 根据判断结果输出答案,我们可以用一个 set 容器来对颜色进行去重,set 容器会帮我们将相同的值去掉。
    代码
    #include
    using namespace std;
    
    const int N = 10010;
    
    struct Edge {
        int a, b;
    }e[N];
    
    int n, m;
    int color[N];
    
    int main()
    {
        cin >> n >> m;
    
        //输入每条边
        for (int i = 0; i < m; i++)    cin >> e[i].a >> e[i].b;
    
        //开始查询
        int k;
        cin >> k;
        while (k--)
        {
            //输入每种染色方案
            for (int i = 0; i < n; i++)  cin >> color[i];
    
            //判断每条边的两个点颜色是否都是不同
            bool success = true;
            for (int i = 0; i < m; i++)
                if (color[e[i].a] == color[e[i].b])
                {
                    success = false;
                    break;
                }
    
            if (success)
            {
                //可以用set来对颜色进行去重
                unordered_set<int> s;
                for (int i = 0; i < n; i++)  s.insert(color[i]);
                printf("%d-coloring\n", s.size());
            }
            else puts("No");
        }
    
        return 0;
    }
    
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  • 原文地址:https://blog.csdn.net/Newin2020/article/details/127589219