题目

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. Afer 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

题目分析

给出图的每条边顶点信息,给出几组顶点集合,判断顶点集合是否是vertex cover(vertex cover指:一个顶点集合,图每条边的顶点至少有一个在这个顶点集合中)

解题思路

算法 1

  1. 定义顶点结构体edge,两个顶点left,right
  2. 定义vector v,存放图每条边的信息
  3. 定义int ves[N],存放每个查询顶点集合中顶点出现次数
  4. 每个顶点集合的判断,需要遍历所有边信息
    • 每条边的两个顶点至少有一个在顶点集合中,满足条件,打印Yes
    • 只要有一条边的两个顶点都不在顶点集合中,不满足条件,打印No

算法2

  1. 定义一个vector数组,数组下标表示顶点,vector中存放的是该顶点所在边的编号
  2. 每次校验一个顶点集合,定义一个int hash[M]数组,下标为图的边,值为边的顶点是否至少有一个在顶点集合中,若边满足条件置为1
  3. 遍历hash[M]数组,是否所有元素都为1,表示每条边至少有一个顶点在顶点集合中,该顶点集合是vertex cover

Code

Code 01(算法1 最优)

#include <iostream>
#include <vector>
using namespace std;
struct edge {
int left,right;
};
int main(int argc,char * argv[]) {
int N,M,K,L,V;
scanf("%d %d", &N,&M);
edge es[M];
for(int i=0; i<M; i++) {
scanf("%d %d",&es[i].left,&es[i].right);
}
scanf("%d",&K);
for(int i=0; i<K; i++) {
scanf("%d", &L);
int ves[N]= {0};
for(int j=0; j<L; j++) {
scanf("%d",&V);
ves[V]++;
}
int j;
for(j=0; j<M; j++) {
if(ves[es[j].left]==0&&ves[es[j].right]==0)break;
}
if(j!=M)printf("No\n");
else printf("Yes\n");
} return 0;
}

Code 02(算法2)

#include <iostream>
#include <vector>
using namespace std;
int main(int argc,char * argv[]) {
int N,M,K,L,V;
scanf("%d %d", &N,&M);
vector<int> es[N];
int f,r;
for(int i=0; i<M; i++) {
scanf("%d %d",&f,&r);
es[f].push_back(i);
es[r].push_back(i);
}
scanf("%d",&K);
for(int i=0; i<K; i++) {
scanf("%d", &L);
int hash[M]= {0};
for(int j=0; j<L; j++) {
scanf("%d",&V);
for(int t=0; t<es[V].size(); t++) {
hash[es[V][t]]=1;
}
}
int j;
for(j=0; j<M; j++) {
if(hash[j]==0) {
break;
}
}
if(j!=M)printf("No\n");
else printf("Yes\n");
} return 0;
}

PAT Advanced 1134 Vertex Cover (25) [hash散列]的更多相关文章

  1. PAT Advanced 1048 Find Coins (25) [Hash散列]

    题目 Eva loves to collect coins from all over the universe, including some other planets like Mars. On ...

  2. PAT Advanced 1084 Broken Keyboard (20) [Hash散列]

    题目 On a broken keyboard, some of the keys are worn out. So when you type some sentences, the charact ...

  3. PAT Advanced 1050 String Subtraction (20) [Hash散列]

    题目 Given two strings S1 and S2, S = S1 – S2 is defined to be the remaining string afer taking all th ...

  4. PAT Advanced 1041 Be Unique (20) [Hash散列]

    题目 Being unique is so important to people on Mars that even their lottery is designed in a unique wa ...

  5. PAT甲级——1134 Vertex Cover (25 分)

    1134 Vertex Cover (考察散列查找,比较水~) 我先在CSDN上发布的该文章,排版稍好https://blog.csdn.net/weixin_44385565/article/det ...

  6. PAT 甲级 1134 Vertex Cover

    https://pintia.cn/problem-sets/994805342720868352/problems/994805346428633088 A vertex cover of a gr ...

  7. 1134. Vertex Cover (25)

    A vertex cover of a graph is a set of vertices such that each edge of the graph is incident to at le ...

  8. PAT Advanced 1154 Vertex Coloring (25) [set,hash]

    题目 A proper vertex coloring is a labeling of the graph's vertices with colors such that no two verti ...

  9. PAT甲题题解-1078. Hashing (25)-hash散列

    二次方探测解决冲突一开始理解错了,难怪一直WA.先寻找key%TSize的index处,如果冲突,那么依此寻找(key+j*j)%TSize的位置,j=1~TSize-1如果都没有空位,则输出'-' ...

随机推荐

  1. JS高级学习笔记(1)- 数据类型及转换规则

    必读: Javascript对象Oject的强制类型转换 JavaScript筑基篇(二)->JavaScript数据类型 聊一聊valueOf和toString 深入理解JavaScript系 ...

  2. 掌握这几点,轻松搞定Essay Cohesion写作

    Cohesion就是衔接,是留学生Essay写作中中一个很重要的评价标准.很多留学生在平时Essay写作中,主体段已经做到了有观点.有例子,字数也不差,但总是被评价为展开不够.说理不清.不好follo ...

  3. 无法启动APK安装也,报异常FileUriExposedException

    无法打开APK安装页,报异常FileUriExposedException, https://juejin.im/entry/58e4643db123db15eb79a902

  4. 细说 OLAP 与 OLTP

    OLAP (Online analytical processing)[联机分析处理] 起源 数据库概念最初源于1962年Kenneth Iverson发表的名为"A Programming ...

  5. 201771010142-张燕 实验一 软件工程准备—<软件工程的初步了解和学习目标>

    实验一 软件工程准备 项目 内容 软件工程 https://www.cnblogs.com/nwnu-daizh/ 软件工程准备要求 https://www.cnblogs.com/nwnu-daiz ...

  6. 关于 with 语句

    class C(object): def __enter__(self): print('jinru') return self def __exit__(self, exc_type, exc_va ...

  7. C++编程学习(五) C++ 存储类

    一.auto 存储类 根据初始化表达式自动推断被声明的变量的类型. auto f=3.14; //double auto s("hello"); //const char* aut ...

  8. 如何搞定Critical Thinking写作?

    受中国传统教育模式与国外一流大学之间的差异的影响,在海外留学的学子们常常会在新的学习生活中面临许多难题,Critical Thinking就是其中之一.国内的教育方法常常以灌输式的教育模式为主,忽略了 ...

  9. linux文件编辑VI命令详解

    vi编辑器是所有Unix及Linux系统下标准的编辑器,它的强大不逊色于任何最新的文本编辑器,这里只是简单地介绍一下它的用法和一小部分指令.由于对Unix及Linux系统的任何版本,vi编辑器是完全相 ...

  10. MySQL表连接原理

    以下文章均来自掘金小测: https://juejin.im/book/5bffcbc9f265da614b11b731/section/5c061b0cf265da612577e0f4 表连接本质: ...