【LG5022】[NOIP2018]旅行

【LG5022】[NOIP2018]旅行

题面

洛谷

题解

首先考虑一棵树的部分分怎么打

直接从根节点开始(dfs)，依次选择编号最小的儿子即可

而此题是一个基环树

怎么办呢？

可以断掉环上的一条边，这样就变为一棵树了

再用上面的方法做即可

(tips) (:)

断环上的边，其实可以直接用(tarjan)把桥求出来

不是桥的就是环上的边

考场上的代码有点乱

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector> 
using namespace std;
inline int gi() {
	register int data = 0, w = 1; 
	register char ch = 0;
	while (ch != '-' && !isdigit(ch)) ch = getchar(); 
	if (ch == '-') w = -1, ch = getchar();
	while (isdigit(ch)) data = data * 10 + ch - '0', ch = getchar();
	return w * data; 
}
#define MAX_N 5005
vector<int> G[MAX_N];
struct Graph { int to, next; } e[MAX_N << 1]; int fir[MAX_N], e_cnt; 
void clearGraph(){ memset(fir, -1, sizeof(fir)); } 
void Add_Edge(int u, int v) { e[e_cnt] = (Graph){v, fir[u]}; fir[u] = e_cnt++; }
struct Edge { 
	int to, id;
	bool operator < (const Edge &rhs) const { return to < rhs.to; } 
} ; 
vector<Edge> rG[MAX_N]; 
int N, M; 
namespace cpp1 {
	void dfs(int x, int f) {
		printf("%d ", x); 
		for (int i = 0, sz = rG[x].size(); i < sz; i++) {
			int v = rG[x][i].to;
			if (v == f) continue;
			dfs(v, x); 
		} 
	} 
	void main() { dfs(1, 0); putchar('
'); } 
}

namespace cpp2 {
	int dfn[MAX_N], low[MAX_N], tim; 
	bool bridge[MAX_N << 1];
	void tarjan(int x, int id) {
		dfn[x] = low[x] = ++tim;
		for (int i = fir[x]; ~i; i = e[i].next) {
			int v = e[i].to;
			if (!dfn[v]) {
				tarjan(v, i), low[x] = min(low[x], low[v]); 
				if (low[v] > dfn[x]) bridge[i] = bridge[i ^ 1] = 1; 
			} else if (i != (id ^ 1)) low[x] = min(low[x], dfn[v]); 
		} 
	}
	int ans[MAX_N], tmp[MAX_N], cnt;
	bool used[MAX_N << 1]; 
	void dfs(int x, int fa) {
		tmp[++cnt] = x;
		for (int i = 0, sz = rG[x].size(); i < sz; i++) {
			int v = rG[x][i].to, id = rG[x][i].id;
			if (v == fa || used[id]) continue;
			dfs(v, x); 
		} 
	} 
	bool check() {
		for (int i = 1; i <= N; i++) {
			if (ans[i] > tmp[i]) return 1;
			else if (ans[i] < tmp[i]) return 0; 
		} 
		return 0; 
	} 
    void main() { 
		tarjan(1, -1);
		for (int i = 1; i <= N; i++) ans[i] = N + 1;
		for (int i = 0; i < e_cnt; i += 2) {
			if (bridge[i]) continue; 
			used[i] = used[i ^ 1] = 1; cnt = 0;
			dfs(1, 0);
			if (check()) for (int j = 1; j <= N; j++) ans[j] = tmp[j]; 
			used[i] = used[i ^ 1] = 0; 
		}
		for (int i = 1; i <= N; i++) printf("%d ", ans[i]);
		putchar('
'); 
	} 
} 
bool used[MAX_N][MAX_N]; 
int main () {
	N = gi(), M = gi(); 
	for (int i = 1; i <= M; i++) {
		int u = gi(), v = gi();
		G[u].push_back(v), G[v].push_back(u); 
	} 
	for (int i = 1; i <= N; i++) sort(G[i].begin(), G[i].end());
	clearGraph(); 
	for (int i = 1; i <= N; i++)
		for (int j = G[i].size() - 1; j >= 0; j--) {
			int v = G[i][j]; if (used[i][v]) continue; 
			Add_Edge(i, v), Add_Edge(v, i); 
			used[i][v] = used[v][i] = 1; 
		} 
	for (int x = 1; x <= N; x++) {
		for (int i = fir[x]; ~i; i = e[i].next) {
			int v = e[i].to;
			rG[x].push_back((Edge){v, i}); 
		} 
	}
	for (int i = 1; i <= N; i++) sort(rG[i].begin(), rG[i].end()); 
	if (N - 1 == M) cpp1::main(); 
	else cpp2::main(); 
	return 0; 
}