zoukankan      html  css  js  c++  java
  • 【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, 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("
    ");
            if(deg[i] % 2 == 1)odd++;
        }
        if(cnt == 1){
            if(odd == 0)printf("Eulerian
    ");
            else if(odd == 2)printf("Semi-Eulerian
    ");
            else printf("Non-Eulerian
    ");
        }else{
            printf("Non-Eulerian
    ");
        }
        return 0;
    }
    
  • 相关阅读:
    人生中对我影响最大的三位老师
    自我介绍
    对我影响较大的三位老师
    自我介绍
    Java入门到精通——基础篇之static关键字
    天猫优惠券面值可以随意修改
    常用的PC/SC接口函数
    批量删除本地指定扩展名文件工具
    算法:C++排列组合
    Java入门到精通——基础篇之面向对象
  • 原文地址:https://www.cnblogs.com/vinnson/p/10845098.html
Copyright © 2011-2022 走看看