Submission #1093368

#	Time	Username	Problem	Language	Result	Execution time	Memory
1093368		asli_bg	Race (IOI11_race)	C++11	0 / 100	6 ms	12120 KiB

race

#pragma GCC optimize("O3,unroll-loops")

#include "race.h"

#include <bits/stdc++.h>
using namespace std;
 
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
 
typedef tree<int,null_type,less<int>,rb_tree_tag,
tree_order_statistics_node_update> inset;
 
//#define int long long
 
#define fi first
#define se second
#define all(x) x.begin(),x.end()
#define pb push_back
 
#define FOR(i,a) for(int i=0;i<(a);i++)
#define FORE(i,a,b) for(int i=(a);i<(b);i++)
#define cont(x) {for(auto el:x) cout<<el<<' ';cout<<endl;}
#define contp(x) {for(auto el:x) cout<<el.fi<<'-'<<el.se<<' ';cout<<endl;}
#define sp <<" "<<
#define mid (l+r)/2
#define endl '\n'
 
#define DEBUG(X) {cout<<#X<<' '<<(X)<<endl;}
 
#define carp(x,y) ((x%MOD)*(y%MOD))%MOD
#define topla(x,y) ((x%MOD)+(y%MOD))%MOD
 
typedef vector<int> vi;
typedef pair<int,int> pii;
typedef vector<pii> vii;
typedef vector<bool> vb;
 
const int MAXN=5e5+5;
const int INF=1e9+7;
const int MOD=1e9+7;

vii adj[MAXN];
int n,k;
int res;

void merge(set<pii>& a,set<pii>& b, int cost){
    if(a.size()<b.size()) swap(a,b);
    //setteki elemanlar--> {dist,edge sayisi}
    //adjdeki elemanlar--> {kom,dist}
    for(auto el:b){
        el.fi+=cost;
        if(k-el.fi<0) break;
        auto it=a.lower_bound({k-el.fi,0});
        int bir=(*it).fi;
        int iki=(*it).se;
        if(bir==k-el.fi){
            res=min(res,iki+el.se+1+(bir!=0));
        }
    }

    for(auto el:b) a.insert({el.fi+cost,el.se+1});
}

set<pii> dfs(int nd,int ata){
    set<pii> s;

    if(adj[nd].size()<=1 and nd!=1){
        s.insert({0,0});
        return s;
    }

    for(auto el:adj[nd]){
        int kom=el.fi;
        int cost=el.se;
        if(kom!=ata){
            auto c=dfs(kom,nd);
            merge(s,c,cost);
        }
    }

    return s;
}

int best_path(int N, int K, int H[][2], int L[]){
    res=INF;
    k=K;
    n=N;
    FOR(i,n-1){
        adj[H[i][0]].pb({H[i][1],L[i]});
        adj[H[i][1]].pb({H[i][0],L[i]});
    }
    dfs(1,0);

    if(res==INF) res=-1;
    return res;
}

• Subtask 1: 0 / 9
• Subtask 2: 0 / 12
• Subtask 3: 0 / 22
• Subtask 4: 0 / 57