CodeForces 780 E 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; }