UVA 10441 - Catenyms（欧拉道路）

UVA 10441 - Catenyms

题目链接

题意：给定一些单词，求拼接起来，字典序最小的，注意这里的字典序为一个个单词比过去，并非一个个字母

思路：欧拉回路。利用并查集判联通，然后欧拉道路判定，最后dfs输出路径

代码：

#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
 
const int N = 30;
 
vector<string> g[N];
vector<string> ans;
 
int t, n, parent[N];
bool used[N][1005];
 
int find(int x) {
	return parent[x] == x ? x : parent[x] = find(parent[x]);
}
 
int vis[N];
int cnt, tot, ru[N], chu[N];
 
void init() {
	memset(ru, 0, sizeof(ru));
	memset(chu, 0, sizeof(chu));
	memset(vis, 0, sizeof(vis));
	memset(used, 0, sizeof(used));
	for (int i = 0; i < 26; i++) {
		g[i].clear();
		parent[i] = i;
	}
	cnt = 1; tot = 0;
	scanf("%d", &n);
	string s;
	while (n--) {
		cin >> s;
		int u = s[0] - 'a', v = s[s.length() - 1] - 'a';
		if (!vis[u]) {vis[u] = 1; tot++;}
		if (!vis[v]) {vis[v] = 1; tot++;}
		ru[v]++; chu[u]++;
		g[u].push_back(s);
		int pu = find(u);
		int pv = find(v);
		if (pu != pv) {
			parent[pu] = pv;
			cnt++;
		}
	}
	for (int i = 0; i < 26; i++)
		sort(g[i].begin(), g[i].end());
}
 
void dfs(int u) {
	for (int i = 0; i < g[u].size(); i++) {
		int v = g[u][i][g[u][i].length() - 1] - 'a';
		if (used[u][i]) continue;
		used[u][i] = 1;
		dfs(v);
		ans.push_back(g[u][i]);
	}
}
 
bool solve() {
	if (cnt != tot) return false;
	int Min = 30;
	int odd = 0, st;
	for (int i = 0; i < 26; i++) {
		if (g[i].size()) Min = min(Min, i);
		if (ru[i] - chu[i] == -1) {
			odd++;
			st = i;
		}
		else if (chu[i] - ru[i] == -1)
			odd++;
		else if (chu[i] != ru[i]) return false;
		if (odd > 2) return false;
	}
	ans.clear();
	if (!odd) dfs(Min);
	else dfs(st);
	for (int i = ans.size() - 1; i > 0; i--)
		cout << ans[i] << ".";
	cout << ans[0] << endl;
	return true;
}
 
int main() {
	scanf("%d", &t);
	while (t--) {
		init();
		if (!solve()) printf("***\n");
	}
	return 0;
}