UVA - 10129 Play on Words（欧拉回路）

题意：将n个单词排成一个序列，保证相邻单词相邻处字母相同。

分析：每个单词看做一条有向边，字母为点，并查集看图是否连通，因为是有向图，所以最多只能有两个点入度不等于出度，且这两个点一个入度比出度大1，一个出度比入度大1

并查集，单词的首字母是尾字母的祖先。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MAXN = 1000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
char s[MAXN];
int fa[30];
int in[30];
int out[30];
int Find(int x){
    return fa[x] = (x == fa[x]) ? x : Find(fa[x]);
}
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        for(int i = 0; i <= 25; ++i){
            fa[i] = i;
        }
        memset(in, 0, sizeof in);
        memset(out, 0, sizeof out);
        int n;
        scanf("%d", &n);
        for(int i = 0; i < n; ++i){
            scanf("%s", s);
            int len = strlen(s);
            int x = s[0] - 'a';
            int y = s[len - 1] - 'a';
            ++in[y];
            ++out[x];
            int tx = Find(x);
            int ty = Find(y);
            fa[ty] = tx;//首字母是尾字母的祖先
        }
        bool ok = true;
        int cnt = 0;
        for(int i = 0; i < 26; ++i){//统计连通块个数
            if((in[i] || out[i]) && fa[i] == i){
                ++cnt;
            }
        }
        if(cnt > 1) ok = false;//图不连通
        int num1 = 0;//入度比出度大1的结点数
        int num2 = 0;//出度比入度大1的结点数
        for(int i = 0; i < 26; ++i){
            if(!ok) break;
            if(in[i] != out[i]){
                if(in[i] - out[i] == 1) ++num1;
                else if(out[i] - in[i] == 1) ++num2;
                else{
                    ok = false;
                    break;
                }
            }
        }
        if(ok){
            if(!((num1 == 0 && num2 == 0) || (num1 == 1 && num2 == 1))) ok = false;
        }
        if(!ok){
            printf("The door cannot be opened.\n");
        }
        else{
            printf("Ordering is possible.\n");
        }
    }
    return 0;
}