Submission #492611

#	Time	Username	Problem	Language	Result	Execution time	Memory
492611		Biaz	Sightseeing (NOI14_sightseeing)	C++17	9 / 25	1186 ms	262148 KiB

sightseeing

#include <bits/stdc++.h>
#define int long long
//#define double long double
#define Nanase_Kurumi_aka_menhera_chan_is_mine ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define pb push_back
#define pi pair<int, int>
#define ALL(i) i.begin(),i.end()
#define gcd(i,j) __gcd(i,j)
#define fi first
#define se second
#define eps 0.00000001
#define ist insert
#define DNE nullptr
//#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
//#pragma GCC optimize("O2")
int max(int x,int y){return x>=y?x:y;}
int min(int x,int y){return x>=y?y:x;}
using namespace std;
typedef int ll;
const int N=500005;
const int M=5000005;
const int MOD=1000000007;//998244353;
const int INF=1000000000000000000;//2147483647;

int n,m,Q;
struct I{
    int a,b,c;
    bool operator<(I const& rh)const{
        return c>rh.c;
    }
}e[M];
int p[N];
int res[N];
vector<int> gr[N];
int boss(int a){return a==p[a]?a:p[a]=boss(p[a]);}
void merge(int a,int b){
    if (gr[a].size()>gr[b].size()){
        p[a]=b;
        for (int &i:gr[a])
            gr[b].pb(i);
    }
    else{
        p[b]=a;
        for (int &i:gr[b])
            gr[a].pb(i);
    }
}
inline void sol(){
    cin >>n>>m>>Q;
    for (int i=0;i<m;i++)
        cin >>e[i].a>>e[i].b>>e[i].c;
    sort(e,e+m);
    for (int i=1;i<=n;i++) p[i]=i,gr[i].pb(i);
    for (int i=0;i<m;i++){
        int x=boss(e[i].a),y=boss(e[i].b);
        if (x==y) continue;
        merge(x,y);
        if (boss(x)==boss(1)){
            for (int &v:gr[x])
                if (res[v]==0)
                    res[v]=e[i].c;
        }
        if (boss(y)==boss(1)){
            for (int &v:gr[y])
                if (res[v]==0)
                    res[v]=e[i].c;
        }
    }
    for (int i=0,t;i<Q;i++){
        cin >>t;
        cout <<res[t]<<'\n';
    }
}
signed main(){
    Nanase_Kurumi_aka_menhera_chan_is_mine
    int _=1;
    //cin >>_;
    while (_--) sol();
    return 0;
}
/*
4 4 2
1 2 10
1 3 30
2 4 20
3 4 5
3
4*/

• Subtask 1: 4 / 4
• Subtask 2: 5 / 5
• Subtask 3: 0 / 6
• Subtask 4: 0 / 10