CodeForces 780 E Underground Lab

Underground Lab

题解:

如果遍历一棵树，我们可以发现最多需要走的步数也不会超过2 * n步。

所以我们选出一棵树，然后遍历一边这颗树。

然后把序列分成k块就好了。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 2e5 + 100;
int pre[N];
int Find(int x){
    if(pre[x] == x) return x;
    return pre[x] = Find(pre[x]);
}
vector<int> ans;
vector<int> c[N];
vector<int> vc[N];
void dfs(int o, int u){
    ans.pb(u);
    for(int v : vc[u]){
        if(v == o) continue;
        dfs(u, v);
        ans.pb(u);
    }
}
int Ac(){
    int n, m, k, u, v;
    scanf("%d%d%d", &n, &m, &k);
    for(int i = 1; i <= n; ++i) pre[i] = i;
    for(int i = 1; i <= m; ++i){
        scanf("%d%d", &u, &v);
        int fu = Find(u), fv = Find(v);
//        cout << fu <<  " with " << fv << endl;
        if(fu == fv) continue;
        pre[fu] = fv;
        vc[u].pb(v); vc[v].pb(u);
    }
    dfs(0, 1);
//    cout << "____" << endl;
    int can = (2*n  + k-1)/k;
    for(int i = 0; i < ans.size(); ++i){
        c[i/can].pb(ans[i]);
    }
    for(int i = 0; i < k; ++i){
        if(c[i].size() == 0) c[i].pb(1);
        printf("%d", c[i].size());
        for(int v : c[i]){
            printf(" %d", v);
        }
        puts("");
    }
    return 0;
}

int main(){
    Ac();
    return 0;
}