#include<iostream>
#include<stdio.h>
#include<string.h>
#include<vector>
using namespace std;
int n,m,a,b;
int degree[1005];
int vis[1005]={0};
vector<int> vec[1005];

int get=0;
void dfs(int st)
{
    vis[st]=1;
    get++;
    for(int i=0;i<vec[st].size();i++)
    {
        
 
int v=vec[st][i];
        if(!vis[v])
        {
            dfs(v);
        }
    }
}
int main()
{
    memset(degree,0,sizeof(degree));

    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d",&a,&b);
        vec[a].push_back(b);
        vec[b].push_back(a);
        degree[a]
 
++;
        degree[b]++;
    }
    int flag=1;
    dfs(1);
    if(get!=n) flag=0;
    for(int i=1;i<=n;i++)
    {
        if(degree[i]%2!=0)
        {
            flag=0;
            break;
        }
        if(degree[i]==0)
        {
            flag=0;
            break;
        }
    }
    if(flag) printf("
 
1");
    else printf("0");
    return 0;
}

思路是把字母看成點，單詞看成邊。不知道該怎麽判斷連通性，看了別人用並查集進行判定

#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <string>
#include <cmath>

using namespace std;
int n,chu[200],ru[200],zifu[200],tot,fa[200];

int find(int x)
{
    if (x == fa[x])
        return x;
    return fa[x] = find(fa[x]);
}

bool check(int c)
{
    for (int i = 1; i <= tot; i++)
        if (zifu[i] == c)
            return true;
    return false;
}

int main()
{
    for (int i = ‘a‘;i <= ‘z‘; i++)
        fa[i] = i;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
    {
        char s[25];
        scanf("%s", s);
        int len = strlen(s);
        if (!check(s[0]))
        zifu[++tot] = s[0];
        if (!check(s[len - 1]))
        zifu[++tot] = s[len - 1];
        ru[s[len - 1]]++;
        chu[s[0]]++;
        int x = find(s[0]);
        int y = find(s[len - 1]);
        fa[x] = y;
    }
    if (n == 1)
        printf("Euler path");
    else
    {
        bool flag1 = 0;
        int ans = 0;
        for (int i = 1; i <= tot; i++)
        {
            if (find(zifu[i]) != fa[zifu[1]])
            {
                flag1 = 1; //不連通
                break;
            }
            int t = abs(chu[zifu[i]] - ru[zifu[i]]);
            if (t >= 2)
            {
                //不是歐拉路徑
                flag1 = 1;
                break;
            }
            if (t == 1)
            {
                ans++;
            }
        }
        if (flag1)
            printf("impossible");
        else
            if (ans==0)
                printf("Euler loop");
            else
                if (ans == 2)
                    printf("Euler path");
                else
                    printf("impossible");
    }
    return 0;
}

兩個歐拉路徑的例題