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  • 1134 Vertex Cover

    A vertex cover of a graph is a set of vertices such that each edge of the graph is incident to at least one vertex of the set. Now given a graph with several vertex sets, you are supposed to tell if each of them is a vertex cover or not.

    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 1), being the total numbers of vertices and the 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 queries. Then K lines of queries follow, each in the format: N​v​​ v[1] v[2]⋯v[N​v​​]. where N​v​​ is the number of vertices in the set, and ['s are the indices of the vertices.

    Output Specification:

    For each query, print in a line Yes if the set is a vertex cover, 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
    5
    4 0 3 8 4
    6 6 1 7 5 4 9
    3 1 8 4
    2 2 8
    7 9 8 7 6 5 4 2
    
     

    Sample Output:

    No
    Yes
    Yes
    No
    No

    题意:

      给出一个图和一个图中部分顶点所构成的集合,判断图中的所有边是不是都和集合中的顶点相关。

    思路:

      利用邻接矩阵来存储图,然后查询与集合中顶点相连的边的条数(注意不要重复)是不是与M(图中边数的总和)相等。

    Code:

     1 #include <bits/stdc++.h>
     2 
     3 using namespace std;
     4 
     5 int main() {
     6     int n, m, k;
     7     cin >> n >> m;
     8 
     9     vector<vector<int> > grap(n + 1);
    10     int v1, v2;
    11     for (int i = 0; i < m; ++i) {
    12         cin >> v1 >> v2;
    13         grap[v1].push_back(v2);
    14         grap[v2].push_back(v1);
    15     }
    16 
    17     cin >> k;
    18     int nv, t, edges;
    19     unordered_set<int> uset;
    20     for (int i = 0; i < k; ++i) {
    21         cin >> nv;
    22         uset.clear();
    23         edges = 0;
    24         for (int j = 0; j < nv; ++j) {
    25             cin >> t;
    26             uset.insert(t);
    27             for (int k : grap[t])
    28                 if (uset.find(k) == uset.end()) edges++;
    29         }
    30         if (edges == m)
    31             cout << "Yes" << endl;
    32         else
    33             cout << "No" << endl;
    34     }
    35     return 0;
    36 }
    永远渴望,大智若愚(stay hungry, stay foolish)
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  • 原文地址:https://www.cnblogs.com/h-hkai/p/12735204.html
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