树的性质和dfs的性质 Codeforces Round #403 (Div. 2, based on Technocup 2017 Finals) E

http://codeforces.com/contest/782/problem/E

题目大意：

有n个节点，m条边，k个人，k个人中每个人都可以从任意起点开始走(2*n)/k步,且这个步数是向上取整的。要求：着k个人要走完所有的节点，且每个人至少走1步。

思路：= =dfs找一棵树，一棵树是n-1条边，所以从树根开始走完所有的，也就只有n*2-2步，所以不会达到上限。md这么简单都没有想到，233

//看看会不会爆int!数组会不会少了一维！
//取物问题一定要小心先手胜利的条件
#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker,"/STACK:102400000,102400000")
#define LL long long
#define ALL(a) a.begin(), a.end()
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define haha printf("haha
")
const int maxn = 2e5 + 10;
int n, m, k;
vector<int> G[maxn];
bool vis[maxn];
vector<int> ans[maxn];
int lim, cnt, rest;

void dfs(int u){
    vis[u] = true;
    ans[cnt].pb(u); rest--;
    if (rest == 0) {cnt++; rest = lim;}
    for (int i = 0; i < G[u].size(); i++){
        int v = G[u][i];
        if (vis[v]) continue;
        dfs(v);
        ans[cnt].pb(u); rest--;
        if (rest == 0) {cnt++; rest = lim;}
    }
}

int main(){
    cin >> n >> m >> k;
    for (int i = 1; i <= m; i++){
        int u, v; scanf("%d%d", &u, &v);
        G[u].pb(v); G[v].pb(u);
    }
    lim = n * 2 / k;
    if ((2 * n) % k) lim++;
    rest = lim, cnt = 1;
    dfs(1);
    if (rest == lim) cnt--;
    for (int i = cnt + 1; i <= k; i++){
        ans[i].pb(1); ans[i].pb(G[1][0]);
    }
    for (int i = 1; i <= k; i++){
        printf("%d ", ans[i].size());
        for (int j = 0; j < ans[i].size(); j++){
            printf("%d ", ans[i][j]);
        }
        cout << endl;
    }
    return 0;
}