【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('\n'); }

}

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('\n');

	}

}

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;

}

巴特西

【LG5022】[NOIP2018]旅行

【LG5022】[NOIP2018]旅行

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