| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 538927 | BossBobster | 경주 (Race) (IOI11_race) | C++17 | 0 ms | 0 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <iostream>
#include <string.h>
#include <random>
#include <fstream>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <iomanip>
#include <algorithm>
#include <math.h>
#include <cmath>
#include <vector>
#include <stack>
#include <queue>
#include <bitset>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <complex>
#include <valarray>
using namespace std;
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
//template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef pair<int, int> pii;
typedef pair<int, string> pis;
typedef pair<int, short> pish;
typedef pair<string, string> pss;
typedef pair<int, char> pic;
typedef pair<pii, int> piii;
typedef pair<double, double> pdd;
typedef pair<float, float> pff;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef pair<ll, ll> pll;
typedef pair<pll, ll> plll;
typedef pair<pll, ld> plld;
typedef pair<int, ll> pil;
typedef pair<ull, ull> pull;
typedef pair<ld, ld> pld;
typedef complex<double> cd;
//#define max(n, m) ((n>m)?n:m)
//#define min(n, m) ((n<m)?n:m)
#define f first
#define s second
#define input() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define eps 1e-9
//#define cin fin
vector<pil> adj[200010];
ll n, k, a, b, root;
ll sz[200010];
ll p[200010];
bitset<200010> vis;
ll dfs1(int cur, int par)
{
sz[cur] = 1;
for(auto nx : adj[cur])
if(nx.f != par && !vis[nx.f])
sz[cur] += dfs1(nx.f, cur);
return sz[cur];
}
int dfs2(int cur, int par, ll n)
{
for(auto nx : adj[cur])
if(nx.f != par && sz[nx.f] > n/2 && !vis[nx.f])
return dfs2(nx.f, cur, n);
return cur;
}
ll ans = 2000000000000000;
ll cnt[1000010];
vector<ll> gud;
void dfs3(int cur, int par, ll d, ll num)
{
if(d > k) return;
ans = min(ans, num+cnt[k-d]);
for(auto nx : adj[cur])
if(nx.f != par && !vis[nx.f])
dfs3(nx.f, cur, d+nx.s, num+1);
}
void dfs4(int cur, int par, ll d, ll num)
{
if(d > k) return;
cnt[d] = min(cnt[d], num);
gud.push_back(d);
for(auto nx : adj[cur])
if(nx.f != par && !vis[nx.f])
dfs4(nx.f, cur, d+nx.s, num+1);
}
void make(int cur, int par)
{
int cent = dfs2(cur, par, dfs1(cur, par));
vis[cent] = 1;
gud.clear();
for(auto it : adj[cent])
if(!vis[it.f])
dfs3(it.f, cent, it.s, 1), dfs4(it.f, cent, it.s, 1);
if(cnt[k])
ans = min(ans, cnt[k]);
for(auto it : gud)
cnt[it] = 1000000000;
for(auto it : adj[cent])
if(!vis[it.f])
make(it.f, cent);
}
int best_path(int N, int K, int H[][2], int L[])
{
for(int i = 0; i < n-1; i ++)
adj[H[i][0]].push_back({H[i][1], L[i]}), adj[H[i][1]].push_back({H[i][0], L[i]});
k = K; n = N;
for(int i = 0; i < 1000010; i ++)
cnt[i] = 1000000000;
make(0, -1);
return ((ans>=1000000000)?-1:(int)ans);
}
int edges[200010][2], lll[200010];
int main()
{
input();
cin >> n >> k;
for(int i = 0; i < n-1; i ++)
cin >> edges[i][0] >> edges[i][1] >> lll[i];
cout << best_path(n, k, edges, lll) << '\n';
}