PAT-1134 Vertex Cover （图的建立 + set容器）

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 104), 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:Nv v[1] v[2] ... v[Nv]
where Nv is the number of vertices in the set, and v[i]'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

题目大意：n个顶点和m条边的图，分别给出m条边的两端顶点，然后对其进行k次查询，每次查询输入一个顶点集合，要求判断这个顶点集合是否能完成顶点覆盖，即图中的每一条边都至少有一个顶点在这个集合中。

主要思路：这道题最关键的就是图的建立，这里图的输入并不是给出的顶点及其邻接点的关系，而是给出所有边的两端顶点，如果仍然用二维矩阵的方法构造，后续的操作很容易就超时。这里用一个二维的vector容器，对于图中的每个点，添加其所有关联的边，用0 ~ m-1代表所有的m条边，图模型就构造好了。接着，对于点覆盖问题，可以转化成判断集合中每个点所关联的边加起来是否等于图的边数m，由于这里边不能重复，自然而然想到set容器，将要查询的集合中每个点的所有关联边加入set，如果数量等于图的边数m，则完成顶点覆盖，否则不能。

#include <iostream>
#include <vector>
#include <set>
using namespace std;
int main(void) {
    int n, m, i, j;
    
    cin >> n >> m;
    vector<vector<int> > edge(n);	
    for (i = 0; i < m; i++) {
        int v, w;
        cin >> v >> w;
        edge[v].push_back(i);
        edge[w].push_back(i);
    }
    
    int k, nv, ver, t;
    set<int> s;
    cin >> k;
    for (i = 0; i < k; i++) {
        cin >> nv;
        for (j = 0; j < nv; j++) {
            cin >> ver;
			for (t = 0; t < edge[ver].size(); t++)
				s.insert(edge[ver][t]);
        }
        if (s.size() == m)  cout << "Yes" << endl;
        else                cout << "No" << endl;
        s.clear();                  //清空set集合
    }
    
    return 0;
}

总结：构造模型时不要定式思维；关于查询，尽量用少的数据去多的数据里查，避免从多的数据往少的数据里查。