제출 #320874

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
320874		zaneyu	악어의 지하 도시 (IOI11_crocodile)	C++14	100 / 100	1382 ms	117572 KiB

crocodile

#include "crocodile.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;
typedef tree<long long, null_type, less_equal<long long>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#pragma GCC optimize("unroll-loops,no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
typedef long long int ll;
#define ld double
#define pii pair<ll,ll>
#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()
const ll maxn=5e3+5;
const ll maxlg=__lg(maxn)+2;
const ll INF64=4e18;
const int INF=0x3f3f3f3f;
const ll MOD=(1e9+7);
const ll MOD2=1000002173;
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)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
ll mult(ll a,ll b){
    return ((a%MOD)*(b%MOD))%MOD;
}
ll mypow(ll a,ll b,ll MOD){
    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;
}
#define ll long long
vector<pii> adj[100005];
int vis[100005];
ll dist[100005];
 
 
int travel_plan(int N, int M, int R[][2], int L[], int K, int P[])
{
  REP(i,M){
    adj[R[i][0]].pb({R[i][1],L[i]});
    adj[R[i][1]].pb({R[i][0],L[i]});
  }
  priority_queue<pii> pq;
  REP(i,K){
    pq.push({0,P[i]});
    pq.push({0,P[i]});
  }
  while(!pq.empty()){
    ll d=-pq.top().first;
    int node=pq.top().second;
    pq.pop();
    if(++vis[node]!=2) continue;
    dist[node]=d;
    for(auto e:adj[node]){
      int child=e.first;
      pq.push({-dist[node]-e.second,child});
    }
  }
  return dist[0];
}

• Subtask 1: 46 / 46
• Subtask 2: 43 / 43
• Subtask 3: 11 / 11