제출 #317802

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
317802		zaneyu	경주 (Race) (IOI11_race)	C++14	0 / 100	22 ms	35584 KiB

race

/*input
3 3
0 1 1
1 2 1
-1

*/
#include "race.h"
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
 
#pragma GCC optimize("unroll-loops,no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
#define ll long long 
#define ld long double
#define pii pair<int,int>
#define piii pair<int,pii>
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(ll i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define MP make_pair
const ll INF64=4e12;
const int INF=0x3f3f3f3f;
const ll MOD=1e9+7;
const ld PI=acos(-1);
const ld eps=1e-9;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
ll mult(ll a,ll b){
    return ((a%MOD)*(b%MOD))%MOD;
}
ll mypow(ll a,ll b){
    if(b<=0) return 1;
    ll res=1LL;
    while(b){
        if(b&1){
            res=(res*a)%MOD;
        }
        a=(a*a)%MOD;
        b>>=1;
    }
    return res;
}
const ll maxn=5e5+5;
const ll maxlg=__lg(maxn)+2;
ll pref[maxn];
int n,k;
int ans=INF;
int rt[maxn];
map<int,int> mp[maxn];
vector<pii> v[maxn];
pii bias[maxn];
void dfs(int u,int p){
    rt[u]=u;
    mp[u][0]=0;
    for(auto xx:v[u]){
        int vv=xx.f;
        if(vv==p) continue;
        dfs(vv,u);
        int ru=rt[u],rv=rt[vv];
        if(sz(mp[ru])<sz(mp[rv])) swap(ru,rv);
        bias[rv].f+=xx.s;
        bias[rv].s++;
        for(auto x:mp[rv]){
            if(mp[ru].count(k-x.f-bias[rv].f-bias[ru].f)){
                MNTO(ans,mp[ru][k-x.f-bias[rv].f-bias[ru].f]+bias[ru].s+bias[rv].s+x.s);
            }
        }
        for(auto x:mp[rv]){
            if(x.f+bias[rv].f-bias[ru].f>=0 and (!mp[ru].count(x.f+bias[rv].f-bias[ru].f) or mp[ru][x.f+bias[rv].f-bias[ru].f]>x.s+bias[rv].s-bias[ru].s)){
                mp[ru][x.f+bias[rv].f-bias[ru].f]=x.s+bias[rv].s-bias[ru].s;
            }
        }
        rt[u]=ru;
        rt[vv]=ru;
        bias[rv]={0,0};
        mp[rv].clear();
    }
}
int best_path(int N, int K, int H[][2], int L[]){
    n=N;
    k=K;
    REP(i,n){
        v[H[i][0]].pb({H[i][1],L[i]});
        v[H[i][1]].pb({H[i][0],L[i]});
    }
    dfs(0,-1);
    if(ans>n) return -1;
    return ans;
}

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