1134. Vertex Cover (25)

最新推荐文章于 2021-09-08 16:00:43 发布

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最新推荐文章于 2021-09-08 16:00:43 发布

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本文链接：https://blog.youkuaiyun.com/s136424/article/details/78161778

本文介绍了一种判断顶点覆盖集的有效算法，并通过一个具体的竞赛题目进行了解析。该算法利用了标志位和遍历的方法来验证给定的顶点集是否能够覆盖图中的所有边。

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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

解题思路：

y条边以下标0~y-1的顺序依次输入，用二维数组（列长为2）将每条边的两个端点编号进行记录。

每次flag初始为1，vis数组元素全部指向0,。输入一组端点编号，将与编号相对的vis数组下标元素设为1。

接着，把y条边（0-y-1）依次遍历。若有一条边的两个端点编号均不属于输入的端点编号，则表明没有完全覆盖。此时，用flag=0做标记，跳出循环。

flag=1，输出Yes，否则，输出No。

我的代码：

#include<stdio.h>
#include<memory.h>
int main()
{
	int x,y,z,i,n,m,a[10001][2],vis[10001];
	scanf("%d%d",&x,&y);
	for(i=0;i<y;i++) scanf("%d%d",&a[i][0],&a[i][1]);
	scanf("%d",&z);
	while(z--)
	{
		scanf("%d",&n);
		memset(vis,0,sizeof(vis));
		int flag=1;
		while(n--)
		{
			scanf("%d",&m);
			vis[m]=1;
		}
		for(i=0;i<y;i++)
		{
			if(vis[a[i][0]]==0 && vis[a[i][1]]==0)
			{
				flag=0;
				break;
			}
		}
		if(flag==0) puts("No");
		else puts("Yes");
	}
	return 0;
}

提交结果：