1134. Vertex Cover (25)

1134. Vertex Cover (25)

时间限制

600 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

题意：给出Nv个顶点，要求每条边都与顶点直接相连

思路：给每条边 编号，用邻接表存储  边的编号，遍历Nv个节点的边，给遇到的边上标记。最后如果有没标记的就是No，反之Yes。

代码：

#include <iostream>
#include <algorithm>
#include <vector>
#include <stdio.h>
using namespace std;
vector <int> num[100000];
int main()
{
	int n, m, k;
	cin >> n >> m;
	int a, b;
	for (int i = 0; i < m; i++)
	{
		cin >> a >> b;
		num[a].push_back(i);  
		num[b].push_back(i);
	}
	cin >> k;
	while (k--)
	{
		int judge = 1;
		cin >> a;
		vector <int> edge (m, 0);
		for (int i = 0; i < a; i++)
		{
			cin >> b;
			for (int j = 0; j < num[b].size(); j++) 
				edge[num[b][j]] = 1;
		}
		for (int i = 0; i < m; i++)
			if (!edge[i]) judge = 0;
		if (judge) cout << "Yes" << endl;
		else cout << "No" << endl;
	}
}

因为题目是 Vertex Cover ,所以做的时候总是从点出发，然后思绪就很混乱。看了别人的代码以后才发现只要数边就行了 _(:з)∠)_

思路历程：

先想到的是用邻接矩阵存储，遍历Nv个节点所在行的值，把连通的地方（1）改成0，最后遍历全图，如果还存在边（1） 则输出No，反之Yes 。接着发现是10^4，想想铁定超时。

进而改用邻接表存储，遍历Nv个节点的边的同时删除边，最后如果还存在边，则为No，反之Yes。写的时候发现：

1：有k组数据要测试，直接删会影响后面处理

2：删除对应顶点的边时，得要搜索，时间复杂度提高

之后就走了邪路，想着用点亮顶点的方法  结果第一组数据就是顶点全亮但少一条边的情况

看题解才知道点亮边就行了……

PS:搜的时候看到另外一种思路，保存所有边的信息，对于每一条边的两个顶点，在Vn顶点的集合中查找。如果两个顶点都不在集合中，说明这条边没有被覆盖

 博客网址： http://blog.csdn.net/akibayashi/article/details/78014749

代码：

#include<iostream>
#include<set>
using namespace std;

int main()
{
	int N, M, K, edges[10002][2];
	cin >> N >> M;
	for (int i = 0; i<M; i++)
	{
		int a, b;
		cin >> a >> b;
		edges[i][0] = a;
		edges[i][1] = b;
	}
	cin >> K;
	for (int i = 0; i<K; i++)
	{
		int n;
		bool flag = true;
		set<int> vset;
		cin >> n;
		for (int j = 0; j<n; j++)
		{
			int input;
			cin >> input;
			vset.insert(input);
		}
		for (int j = 0; j<M; j++)
		{
			if (vset.find(edges[j][0]) == vset.end() && vset.find(edges[j][1]) == vset.end())
			{
				flag = false;
				break;
			}
		}
		if (flag)
			cout << "Yes" << endl;
		else
			cout << "No" << endl;
	}

	return 0;
}