PAT 甲级 1126 Eulerian Path

https://pintia.cn/problem-sets/994805342720868352/problems/994805349851185152

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

 

代码：

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e5 + 10;
int N, M;
vector<int> v[maxn];
int vis[maxn];
int step = 0;

void dfs(int st) {
    vis[st] = 1;
    step ++;
    for(int i = 0; i < v[st].size(); i ++) {
        if(vis[v[st][i]] == 0)
            dfs(v[st][i]);
    }
}

int main() {
    scanf("%d%d", &N, &M);
    memset(vis, 0, sizeof(vis));
    while(M --) {
        int a, b;
        scanf("%d%d", &a, &b);
        v[a].push_back(b);
        v[b].push_back(a);
    }

    int cnt = 0;
    for(int i = 1; i <= N; i ++) {
        printf("%d", v[i].size());
        printf("%s", i != N ? " " : "\n");
        if(v[i].size() % 2 == 0) cnt ++;
    }

    dfs(1);
    if(step != N) printf("Non-Eulerian\n");
    else {
        if(cnt == N) printf("Eulerian\n");
        else if(cnt == N - 2) printf("Semi-Eulerian\n");
        else printf("Non-Eulerian\n");
    }
    return 0;
}　　

题目意思是判断每个点的度 都是偶数的话输出 Eulerian 只有两个点的度是奇数其余都是偶数的话输出 Semi-Eulerian 否则输出 Non-Eulerian 要注意 dfs 判断是不是一个连通图

新年快乐 不好的事情就留在上一年吧 今年一定要有不一样