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 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

——————————————————————————————

题目的意思是给出n条边的关系，输出每个点的度和判断是欧拉回路还是欧拉路径还是什么都不是

思路：开数组保存每个点的度，先并查集找出是否是一个连通图，再找出度为奇数的点的数量没如果没有则为欧拉回路，如果是2个则为欧拉路径，否则什么都不是

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<set>
using namespace std;
#define LL long long
const int inf=0x3f3f3f3f;

int cnt[10005];
int pre[10005];
void init()
{
    for(int i=0; i<10005; i++)
        pre[i]=i;
}

int fin(int x)
{
    return x==pre[x]?x:pre[x]=fin(pre[x]);
}

int main()
{
    int m,n,u,v;
    scanf("%d%d",&m,&n);
    memset(cnt,0,sizeof cnt);
    init();
    for(int i=0; i<n; i++)
    {
        scanf("%d%d",&u,&v);
        int a=fin(u);
        int b=fin(v);
        if(a!=b)
            pre[a]=b;
        cnt[u]++,cnt[v]++;
    }
    int ans=0;
    int q=0;
    int flag=0;
    for(int i=1; i<=m; i++)
    {
        if(cnt[i]%2) ans++;
        if(pre[i]==i) flag++;
        if(q++)
            printf(" ");
        printf("%d",cnt[i]);
    }
    printf("\n");
    if(flag==1)
        printf("%s\n",ans==0?"Eulerian":(ans<=2)?"Semi-Eulerian":"Non-Eulerian");
    else
        printf("Non-Eulerian\n");
    return 0;
}