PAT-1134 Vertex Cover (图的建立 + set容器)
2017-09-18 09:53
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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。
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。
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); } 4000 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。
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