【刷题-PAT】A1126 Eulerian Path (25 分)
1126 Eulerian Path (25 分)
In graph theory, an Eulerian path is a path in a graph which visits every edge exactly once. Similarly, an Eulerian circuit is an Eulerian path which starts and ends on the same vertex. They were first discussed by Leonhard Euler while solving the famous Seven Bridges of Konigsberg problem in 1736. It has been proven that connected graphs with all vertices of even degree have an Eulerian circuit, and such graphs are called Eulerian. If there are exactly two vertices of odd degree, all Eulerian paths start at one of them and end at the other. A graph that has an Eulerian path but not an Eulerian circuit is called semi-Eulerian. (Cited from https://en.wikipedia.org/wiki/Eulerian_path)
Given an undirected graph, you are supposed to tell if it is Eulerian, semi-Eulerian, or non-Eulerian.
Input Specification:
Each input file contains one test case. Each case starts with a line containing 2 numbers N (≤ 500), and M, which are the total number of vertices, and the number of edges, respectively. Then M lines follow, each describes an edge by giving the two ends of the edge (the vertices are numbered from 1 to N).
Output Specification:
For each test case, first print in a line the degrees of the vertices in ascending order of their indices. Then in the next line print your conclusion about the graph -- either
Eulerian
,Semi-Eulerian
, orNon-Eulerian
. Note that all the numbers in the first line must be separated by exactly 1 space, and there must be no extra space at the beginning or the end of the line.Sample Input 1:
7 12
5 7
1 2
1 3
2 3
2 4
3 4
5 2
7 6
6 3
4 5
6 4
5 6
Sample Output 1:
2 4 4 4 4 4 2
Eulerian
Sample Input 2:
6 10
1 2
1 3
2 3
2 4
3 4
5 2
6 3
4 5
6 4
5 6
Sample Output 2:
2 4 4 4 3 3
Semi-Eulerian
Sample Input 3:
5 8
1 2
2 5
5 4
4 1
1 3
3 2
3 4
5 3
Sample Output 3:
3 3 4 3 3
Non-Eulerian
分析:注意定义中的path,一个path一定要覆盖所有的节点,也即图要连通
#include<iostream>
#include<cstdio>
#include<vector>
#include<unordered_map>
#include<string>
#include<set>
#include<algorithm>
#include<cmath>
using namespace std;
const int nmax = 510;
int fath[nmax];
void init(){
for(int i = 0; i < nmax; ++i)fath[i] = i;
}
int findF(int x){
int a = x;
while(x != fath[x])x = fath[x];
while(a != fath[a]){
int temp = fath[a];
fath[a] = x;
a = temp;
}
return x;
}
void Union(int a, int b){
int fa = findF(a), fb = findF(b);
if(fa != fb)fath[fa] = fb;
}
bool isRoot[nmax] = {false};
int main(){
#ifdef ONLINE_JUDGE
#else
freopen("input.txt", "r", stdin);
#endif
init();
int n, m;
scanf("%d%d", &n, &m);
int deg[n + 1] = {0};
for(int i = 0; i < m; ++i){
int u, v;
scanf("%d%d", &u, &v);
deg[u]++;
deg[v]++;
Union(u, v);
}
for(int i = 1; i <= n; ++i)isRoot[findF(i)] = true;
int cnt = 0;
for(int i = 1; i <= n; ++i)if(isRoot[i])cnt++;
int odd = 0;
for(int i = 1; i <= n; ++i){
printf("%d", deg[i]);
if(i < n)printf(" ");
else printf("\n");
if(deg[i] % 2 == 1)odd++;
}
if(cnt == 1){
if(odd == 0)printf("Eulerian\n");
else if(odd == 2)printf("Semi-Eulerian\n");
else printf("Non-Eulerian\n");
}else{
printf("Non-Eulerian\n");
}
return 0;
}
【刷题-PAT】A1126 Eulerian Path (25 分)的更多相关文章
- PAT A1126 Eulerian Path (25 分)——连通图,入度
In graph theory, an Eulerian path is a path in a graph which visits every edge exactly once. Similar ...
- PAT甲题题解-1126. Eulerian Path (25)-欧拉回路+并查集判断图的连通性
题目已经告诉如何判断欧拉回路了,剩下的有一点要注意,可能图本身并不连通. 所以这里用并查集来判断图的联通性. #include <iostream> #include <cstdio ...
- PAT甲级 1126. Eulerian Path (25)
1126. Eulerian Path (25) 时间限制 300 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue In grap ...
- PTA PAT排名汇总(25 分)
PAT排名汇总(25 分) 计算机程序设计能力考试(Programming Ability Test,简称PAT)旨在通过统一组织的在线考试及自动评测方法客观地评判考生的算法设计与程序设计实现能力,科 ...
- 【刷题-PAT】A1114 Family Property (25 分)
1114 Family Property (25 分) This time, you are supposed to help us collect the data for family-owned ...
- 【刷题-PAT】A1101 Quick Sort (25 分)
1101 Quick Sort (25 分) There is a classical process named partition in the famous quick sort algorit ...
- PAT甲级——A1126 Eulerian Path【30】
In graph theory, an Eulerian path is a path in a graph which visits every edge exactly once. Similar ...
- PAT 甲级 1029 Median (25 分)(思维题,找两个队列的中位数,没想到)*
1029 Median (25 分) Given an increasing sequence S of N integers, the median is the number at the m ...
- PAT 1126 Eulerian Path[欧拉路][比较]
1126 Eulerian Path (25 分) In graph theory, an Eulerian path is a path in a graph which visits every ...
随机推荐
- AT2664 [AGC017A] Biscuits 题解
Content 有一个长度为 \(n\) 的数列 \(a\).你希望从中选出一些数,使得这些数的和对 \(2\) 取模后的结果为 \(P\).求方案数. 数据范围:\(1\leqslant n\leq ...
- 点击DIV触发其他元素的点击事件(案例:点击type="radio" 的input 标签外层DIV,触发内部单选点击选中事件)
方法一: 直接用找到对应dom元素调用.click()方法 $('.user_content').click(function(){ $(this).children()[0].click(); // ...
- 我写了个IDEA开源插件,vo2dto 一键生成对象转换
让人头疼的对象转换 头炸,po2vo.vo2do.do2dto,一堆对象属性,取出来塞进来.要不是为了 DDD 架构下的各个分层防腐,真想一竿子怼下去. 那上 BeanUtils.copyProper ...
- 【LeetCode】1160. Find Words That Can Be Formed by Characters 解题报告(C++)
作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 字典统计 日期 题目地址:https://leetco ...
- 【LeetCode】295. Find Median from Data Stream 解题报告(C++)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 大根堆+小根堆 日期 题目地址:https://le ...
- Lucky Substrings
而在26以内且属于fibonacci数列的数为1,2,3,5,8,13,21时间限制:10000ms 单点时限:1000ms 内存限制:256MB 描述 A string s is LUCKY if ...
- Nginx应用场景配置
Nginx应用全入门 基础回顾 Nginx是什么? Nginx是一个高性能的HTTP和反向代理web服务器,特点是内存少,并发能力强 Nginx能做什么 Http服务器(Web服务器) 反向代理服务器 ...
- ELK集中化日志解决方案——看这一篇全搞定
一.前言 在软件发开技术管理里有两个永恒经典的问题,适合我们初到一家软件企业或一家公司的科技团队,来判断自己该从哪里入手帮助整个团队提升科技水平和产能.问题一是"在我们团队里,只涉及一行代码 ...
- CS5268替代AG9321MCQ 替代AG9321方案 TYPEC转HDMI多功能拓展坞
台湾安格AG9321MCQ是一款TYPEC拓展坞产品方案,他集中了TYPEC 转HDMI VGA PD3.0快充 QC3.0数据传输 I2S接口的音频DAC输出以及可以各种读卡器功能. Caps ...
- Java EE数据持久化框架 • 【第4章 MyBatis动态SQL】
全部章节 >>>> 本章目录 4.1 MyBatis动态标签 4.1.1 MyBatis动态标签介绍 4.1.2 < if >标签 4.1.3 update语 ...