CF - 782E. Underground Lab - 思维+dfs

1.题目描述：

E. Underground Lab time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

The evil Bumbershoot corporation produces clones for gruesome experiments in a vast underground lab. On one occasion, the corp cloned a boy Andryusha who was smarter than his comrades. Immediately Andryusha understood that something fishy was going on there. He rallied fellow clones to go on a feud against the evil corp, and they set out to find an exit from the lab. The corp had to reduce to destroy the lab complex.

The lab can be pictured as a connected graph with n vertices and m edges. k clones of Andryusha start looking for an exit in some of the vertices. Each clone can traverse any edge once per second. Any number of clones are allowed to be at any vertex simultaneously. Each clone is allowed to stop looking at any time moment, but he must look at his starting vertex at least. The exit can be located at any vertex of the lab, hence each vertex must be visited by at least one clone.

Each clone can visit at most  vertices before the lab explodes.

Your task is to choose starting vertices and searching routes for the clones. Each route can have at most  vertices.

Input

The first line contains three integers n, m, and k (1 ≤ n ≤ 2·105, n - 1 ≤ m ≤ 2·105, 1 ≤ k ≤ n) — the number of vertices and edges in the lab, and the number of clones.

Each of the next m lines contains two integers xi and yi (1 ≤ xi, yi ≤ n) — indices of vertices connected by the respective edge. The graph is allowed to have self-loops and multiple edges.

The graph is guaranteed to be connected.

Output

You should print k lines. i-th of these lines must start with an integer ci () — the number of vertices visited by i-th clone, followed by ci integers — indices of vertices visited by this clone in the order of visiting. You have to print each vertex every time it is visited, regardless if it was visited earlier or not.

It is guaranteed that a valid answer exists.

Examples input 3 2 1 2 1 3 1 output 3 2 1 3 input 5 4 2 1 2 1 3 1 4 1 5 output 3 2 1 3 3 4 1 5 Note

In the first sample case there is only one clone who may visit vertices in order (2, 1, 3), which fits the constraint of 6 vertices per clone.

In the second sample case the two clones can visited vertices in order (2, 1, 3) and (4, 1, 5), which fits the constraint of 5 vertices per clone.

2.题意概述：

给你一个无向图n个顶点m条边，k个克隆人，他们最多行走(2 * n) / k向上取整个步数，要你安排具体每个机器人行走的方案，题目保证有解

3.解题思路：

考虑DFS+回溯的话，发现正好最多走2*n条边，而k个机器人 ⌈2nk⌉×k≥2n 可以总共遍历点大于 2n。那么直接爆搜一遍记录路径就行。然后注意分配当k特别大时候，剩余的机器人只要停留在原地（即1 1）就行。

4.AC代码：

#include <bits/stdc++.h> #define INF 0x3f3f3f3f #define maxn 200100 #define N 1111 #define eps 1e-6 #define pi acos(-1.0) #define e exp(1.0) using namespace std; const int mod = 1e9 + 7; typedef long long ll; typedef unsigned long long ull; vector<int> G[maxn]; bool vis[maxn]; int path[maxn * 2], cnt; void dfs(int from) { vis[from] = 1; path[++cnt] = from; for (int i = 0; i < (int)G[from].size(); i++) { int to = G[from][i]; if (!vis[to]) { dfs(to); path[++cnt] = from; //相当于回溯 } } } int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); freopen("out.txt", "w", stdout); long _begin_time = clock(); #endif int n, m, k; while (~scanf("%d%d%d", &n, &m, &k)) { cnt = 0; memset(vis, 0, sizeof(vis)); for (int i = 1; i <= n; i++) G[i].clear(); for (int i = 0; i < m; i++) { int u, v; scanf("%d%d", &u, &v); G[u].push_back(v); G[v].push_back(u); } dfs(1); int maxC = (2 * n + k - 1) / k; for (int i = 0; i < k; i++) { int c = min(maxC, cnt); if (c) { printf("%d", c); for (int j = 0; j < c && cnt; j++) printf(" %d", path[cnt--]); puts(""); } else puts("1 1"); } } #ifndef ONLINE_JUDGE long _end_time = clock(); printf("time = %ld ms.", _end_time - _begin_time); #endif return 0; }