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  • pat 1154 Vertex Coloring (图的简单模拟)

    1154 Vertex Coloring (25分)

    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 10​4​​), 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

    思路

    本题就是每个点有不同颜色, 题意需要找到一个特殊结构满足相邻边的颜色不能相同,如果存在这种情况,就输出这个图结构一共有多少种颜色,如果相邻边的
    颜色相同,则输出No
    本题邻接矩阵容易超时,因此改用邻接表
    直接将每个点的邻接点都判断一次, 如果存在相同颜色的点就输出no

    代码

    #include<bits/stdc++.h>
    using namespace std;
    const int maxsize = 1e4 + 5;
    //int edge[maxsize][maxsize];
    vector<vector<int> > edge;
    int main()
    {
        int n, m;
        scanf("%d%d", &n, &m);
        edge.resize(n + 1);
        int from, to;
        for(int i = 0; i < m; i++) {
            scanf("%d%d", &from, &to);
            edge[from].push_back(to);
            edge[to].push_back(from);
        }
        int k, color[n];
        scanf("%d", &k);
        set<int> s;
        while(k--) {
            s.clear();
            for(int i = 0; i < n; i++) {
                scanf("%d", &color[i]);
                s.insert(color[i]);
            }
            bool isColor = true;
            for(int i = 0; i < n && isColor; i++) {
                for(int j = 0; j < edge[i].size(); j++) {
                    if(color[i] == color[edge[i][j]]) {
                        isColor = false;
                    }
                }
            }
            if(isColor)
                printf("%d-coloring
    ", s.size());
            else
                printf("No
    ");
        }
        return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/csyxdh/p/12483945.html
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