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, or Non-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

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. 【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 ...

  6. 【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 ...

  7. 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 ...

  8. 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 ...

  9. 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

  1. 快速上手FastJSON

    前言 作为一名后端开发而言肯定会接触数据,把数据提供给前端或者把数据存储起来,目前比较火热的传输格式是json,给前端传json是再常见不过啦,甚至是往db里面直接存入json. 在java层面来说j ...

  2. JAVA中CountDownLatch的简单示例

    public static void main(String[] args) throws InterruptedException { CountDownLatch latch =new Count ...

  3. JS设置网站所有字体变为繁体字

    引入chinese.js var zh_default='n';var zh_choose='t';var zh_expires=7;var zh_class='zh_click';var zh_st ...

  4. 【LeetCode】111. Minimum Depth of Binary Tree 解题报告(Python)

    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 DFS BFS 日期 [LeetCode] 题目地址 ...

  5. 【LeetCode】721. Accounts Merge 解题报告(Python)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址: https://leetcode.com/problems/accounts ...

  6. Harry Potter and the Hide Story(hdu3988)

    Harry Potter and the Hide Story Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 ...

  7. 1021 - Painful Bases

    1021 - Painful Bases   PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Limit: 32 MB As ...

  8. TensorFlow.NET机器学习入门【8】采用GPU进行学习

    随着网络越来约复杂,训练难度越来越大,有条件的可以采用GPU进行学习.本文介绍如何在GPU环境下使用TensorFlow.NET. TensorFlow.NET使用GPU非常的简单,代码不用做任何修改 ...

  9. [数学]高数部分-Part IV 一元函数积分学

    Part IV 一元函数积分学 回到总目录 Part IV 一元函数积分学 不定积分定义 定积分定义 不定积分与定积分的几何意义 牛顿-莱布尼兹公式 / N-L 公式 基本积分公式 点火公式(华里士公 ...

  10. Vue.js高效前端开发 • 【Ant Design of Vue框架基础】

    全部章节 >>>> 文章目录 一.Ant Design of Vue框架 1.Ant Design介绍 2.Ant Design of Vue安装 3.Ant Design o ...