洛谷P5022 旅行 题解 去环/搜索

题目链接：https://www.luogu.org/problem/P5022
这道题目一开始看的时候没有思路，但是看到数据范围里面有一个：
(m = n-1) 或 (m = n) ，一下子有了思路。
当 (m = n-1) 时，这就是一棵树，以1为根节点进行搜索，每次优先访问编号小的点即可。
当 (m = n) 时，可知只有一个环，找到环中对应的所有边，然后遍历每一条环中的边，假设删除它，然后就变成了一棵树。
时间复杂度为：(O(n^2)) 。
实现代码如下：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 5050;
struct Node {
	int u, v, id;
	Node() {}
	Node(int _u, int _v, int _id) { u = _u; v = _v; id = _id; }
} edge[maxn*2], loop_edge[maxn];
vector<Node> g[maxn];
int n, m, p[maxn], pe[maxn], cnt, depth[maxn];
bool vis[maxn], vise[maxn*2], marked[maxn*2];
int ans_seq[maxn], tmp_seq[maxn], ans_cnt, tmp_cnt;
bool cmp(Node a, Node b) {
	return a.v < b.v;
}

void find_loop(int u, int pre, int d) {
	vis[u] = true;
	depth[u] = d;
	int sz = g[u].size();
	for (int i = 0; i < sz; i ++) {
		int v = g[u][i].v, id = g[u][i].id;
		if (v == pre) continue;
		if (!vis[v]) {
			vise[id] = vise[id^1] = true;
			p[v] = u;
			pe[v] = id;
			find_loop(v, u, d+1);
		}
	}
}

void select_loop_edges(int u, int v, int id) {
	loop_edge[cnt++] = Node(u, v, id);
	while (u != v) {
		if (depth[u] > depth[v]) {
			loop_edge[cnt++] = Node(p[u], u, pe[u]);
			u = p[u];
		}
		else if (depth[u] < depth[v]) {
			loop_edge[cnt++] = Node(p[v], v, pe[v]);
			v = p[v];
		}
		else {
			loop_edge[cnt++] = Node(p[u], u, pe[u]);
			u = p[u];
			loop_edge[cnt++] = Node(p[v], v, pe[v]);
			v = p[v];
		}
	}
}

void dfs(int u, int pre) {
	tmp_seq[tmp_cnt++] = u;
	int sz = g[u].size();
	for (int i = 0; i < sz; i ++) {
		int v = g[u][i].v, id = g[u][i].id;
		if (v == pre || marked[id] || marked[id^1]) continue;
		dfs(v, u);
	}
}

void final_check() {
	bool flag = false;
	if (ans_seq[0]) {
		for (int i = 0; i < n; i ++) {
			if (ans_seq[i] > tmp_seq[i]) { flag = true; break; }
			else if (ans_seq[i] < tmp_seq[i]) break;
		}
	}
	else 
		flag = true;
	if (flag) for (int i = 0; i < n; i ++) ans_seq[i] = tmp_seq[i];
}

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 0; i < m; i ++) {
		int u, v, w;
		scanf("%d%d", &u, &v);
		g[u].push_back(Node(u, v, i*2));
		g[v].push_back(Node(v, u, i*2+1));
		edge[i*2] = Node(u, v, i*2);
		edge[i*2+1] = Node(v, u, i*2+1);
	}
	for (int i = 1; i <= n; i ++) sort(g[i].begin(), g[i].end(), cmp);
	if (m == n) {
		find_loop(1, -1, 1);
		for (int i = 0; i < 2*m; i += 2) {
			if (!vise[i]) {
				int u = edge[i].u, v = edge[i].v;
				select_loop_edges(u, v, i);
				break;
			}
		}
		for (int i = 0; i < cnt; i ++) {
			marked[loop_edge[i].id] = marked[loop_edge[i].id^1] = true;
			tmp_cnt = 0;
			dfs(1, -1);
			final_check(); 
			marked[loop_edge[i].id] = marked[loop_edge[i].id^1] = false;
		}
	}
	else {	// m == n-1
		dfs(1, -1);
		final_check();
	}
	for (int i = 0; i < n; i ++) {
		if (i) putchar(' ');
		printf("%d", ans_seq[i]);
	}
	puts("");
	return 0;
}