Submission #360267

#TimeUsernameProblemLanguageResultExecution timeMemory
360267ksmzzang2003Race (IOI11_race)C++17
100 / 100
532 ms31584 KiB
#include "race.h" #include <bits/stdc++.h> using namespace std; #define fi first #define se second #define eb emplace_back #define pb pop_back #define em emplace #define sz(x) ((ll)(x).size()) #define all(x) (x).begin(), (x).end() #define mp make_pair #define upmin(x,y) x=min(x,y) #define upmax(x,y) x=max(x,y) typedef long long ll; typedef pair <int,int> pii; int inf = INT_MAX; ll llinf = LLONG_MAX; int N,K,L[200003]; int par[200003],sz[200003]; vector <pii> adj[200003]; int check[1000003]; bool del[200003]; int ans = 1987654321; vector <pii> lazy; vector <int> usecheck; int dfs1(int now,int prev) { sz[now] =0; for(auto it : adj[now]) { int next = it.first; if(next == prev) continue; if(del[next]) continue; sz[now] += dfs1(next,now); } return ++sz[now]; } int find_cen(int now,int prev,int root) { for(auto it : adj[now]) { int next = it.fi; if(next == prev) continue; if(del[next]) continue; if(sz[next]*2>sz[root]) return find_cen(next,now,root); } return now; } int dfs2(int now,int prev,int sum,int depth,int maxans=1987654321) { if(depth>=maxans) return 1987654321; int ret = 1987654321; if(sum>K) return 1987654321; if(check[K-sum]!=1987654321) upmin(ret,depth + check[K-sum]); lazy.eb(sum,depth); for(auto it : adj[now]) { int next = it.fi; int cost = it.se; if(next == prev) continue; if(del[next]) continue; upmin(ret,dfs2(next,now,sum+cost,depth+1,ret)); if(sum==0) { for(pii it:lazy) upmin(check[it.first],it.second),usecheck.eb(it.first); lazy.clear(); } } return ret; } void g(int now,int prev) { dfs1(now,-1); int cen = find_cen(now,-1,now); for(int it:usecheck) check[it] = 1987654321; usecheck.clear(); check[0] = 0; upmin(ans,dfs2(cen,-1,0,0)); del[cen] = true; for(auto it : adj[cen]) { if(del[it.fi]) continue; g(it.fi,now); } } int best_path(int n, int k, int h[][2], int l[]) { fill(check,check+1000003,1987654321); N= n ; K = k; for(int i=0;i<n-1;i++) { adj[h[i][0]].eb(pii(h[i][1],l[i])); adj[h[i][1]].eb(pii(h[i][0],l[i])); } g(1,-1); return (ans>=1987654300)?-1:ans; }
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