HDU 1116 Play on Words 解题报告 并查集 欧拉通路

HDU 1116 Play on Words 解题报告

解题思路：这道题的欧拉图是有向的，对于每个点都要统计入度和出度，判断时对于所有点，只有一个点入度比出度大1，只有一个点出度比入度大1，其他点入度出度必须相等。用并查集连接点。

#include<iostream>
#include<math.h>
#include<iomanip>
#include<algorithm>
#include<iostream>
#include<math.h>
#include<iomanip>
#include<algorithm>
#include<queue>
#include<cstring>
#include<string>
#include<map>
#include<stack>
#include<stdio.h>
#include<cstdio>
#include<stdlib.h>
#include<fstream>
#include<iomanip>
#pragma warning(disable:4996)
#define INF 0x3f3f3f3f
#define ll long long
#define PI acos(-1.0)
const int N = 200010;
const int maxn = 1e9;
using namespace std;
int n, len;
char a[1005];
int in[27];
int out[27];
int pre[27];
int visit[27];
int find(int x)
{
	if (x != pre[x])
		pre[x] = find(pre[x]);
	return pre[x];
}
void union_set(int a, int b)
{
	int pa = find(a);
	int pb = find(b);
	pre[pb] = pa;
	return;
}
void solve()
{
	scanf("%d", &n);
	memset(in, 0, sizeof(in));
	memset(out, 0, sizeof(out));
	memset(visit, 0, sizeof(visit));
	for (int i = 0; i < 27; i++)
	{
		pre[i] = i;
	}
	for (int i = 0; i < n; i++)
	{
		scanf("%s", &a);
		len = strlen(a);
		int j = a[0] - 'a';
		int k = a[len - 1] - 'a';
		out[j]++;
		in[k]++;
		visit[j] = 1;
		visit[k] = 1;
		union_set(j, k);
		memset(a, 0, sizeof(a));
	}
	int cnt = 0;
	int a = 0, b = 0;
	for (int i = 0; i < 27; i++)
	{
		if (visit[i])
		{
			if (in[i] - out[i] == 1)
			{
				a++;
			}
			if (out[i] - in[i] == 1)
			{
				b++;
			}
			if (out[i] - in[i] != 1&& out[i] - in[i] != -1&& out[i] - in[i] != 0)
			{
				printf("The door cannot be opened.\n");
				return;
			}
			if (pre[i] == i)
			{
				cnt++;
			}
			if (cnt > 1)
			{
				printf("The door cannot be opened.\n");
				return;
			}
		}
	}
	if (a == b && a <= 1)
		printf("Ordering is possible.\n");
	else
		printf("The door cannot be opened.\n");
	return;
}
int main()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		solve();
	}
	return 0;
}