Submission #992523

#	Time	Username	Problem	Language	Result	Execution time	Memory
992523		PedroBigMan	Islands (IOI08_islands)	C++14	0 / 100	2071 ms	98236 KiB

islands

#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#pragma GCC optimize("Ofast")
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <string>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <queue>
#include <deque>
#include <list>
#include <iomanip>
#include <stdlib.h>
#include <time.h>
#include <cstring>
using namespace std;
typedef int ll;
typedef long long int lld;
typedef unsigned long long int ull;
typedef long double ld;
#define REP(i,a,b) for(ll i=(ll) a; i<(ll) b; i++)
#define pb push_back
#define mp make_pair
#define pl pair<ll,ll>
#define ff first
#define ss second
#define whole(x) x.begin(),x.end()
#define DEBUG(i) cout<<"Pedro "<<i<<endl
#define INF 1000000000000000000LL
#define EPS ((ld)0.00000000001)
#define pi ((ld)3.141592653589793)
#define VV(vvvv,NNNN,xxxx); REP(iiiii,0,NNNN) {vvvv.pb(xxxx);}
ll mod=1000000007;

template<class A=ll> 
void Out(vector<A> a) {REP(i,0,a.size()) {cout<<a[i]<<" ";} cout<<endl;}

template<class A=ll>
void In(vector<A> &a, ll N) {A cur; REP(i,0,N) {cin>>cur; a.pb(cur);}} 

vector<ll> f, w; vector<bool> visited;
vector<vector<ll> > adj; ll nei;
lld bestpath;
vector<lld> rm; vector<lld> ps;

vector<ll> Floyd(ll s) //Floyd's Algorithm, O(N) time, O(1) memory, return <element of cycle,length od cycle>
{
    ll a=f[s]; ll b=f[f[s]];
    while(a!=b) {a=f[a]; b=f[f[b]];}
    vector<ll> ans; ans.pb(a); b=f[a];
    while(b!=a) {ans.pb(b); b=f[b];}
    return ans;
}

lld DFS(ll s)
{
    visited[s]=true; lld ans = 0; lld thisval;
    REP(i,0,adj[s].size())
    {
        nei = adj[s][i];
        if(visited[nei]) {continue;}
        thisval = (lld)(w[nei]+DFS(nei));
        bestpath = max(bestpath,ans+thisval);
        ans = max(ans,thisval);
    }
    return ans;
}

void Rolling_Max(ll M, ll L) //ans[i]=j means aj<ai, j<i and j is the largest that fullfills these conditions. If no j exists then a[i]=-1. 
{
	deque<ll> chain; 
    REP(i,0,L)
    {
        while(chain.size()>0 && ps[i]>=ps[chain[chain.size()-1]]) {chain.pop_back();}
        chain.pb(i);
        if(chain.front()==i-M) {chain.pop_front();}
        rm[i]=ps[chain.front()];
    }	
} 


int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
	cout.precision(20);
	ll N; cin>>N; f = vector<ll>(N,-1); w = vector<ll>(N,-1); visited = vector<bool>(N,false); adj=vector<vector<ll> >(N,vector<ll>());
    REP(i,0,N) {cin>>f[i]>>w[i]; f[i]--; adj[i].pb(f[i]); adj[f[i]].pb(i);}
    lld ans = 0LL; ll node; ll C; rm = vector<lld>(2*N,0); ps = vector<lld>(2*N,0);
    REP(s,0,N)
    {
        if(visited[s]) {continue;}
        lld bestval = 0; lld thisval;
        vector<ll> cycle = Floyd(s); C = cycle.size();
        continue;
        REP(i,0,cycle.size()) {node=cycle[i]; visited[node]=true;}
        vector<lld> val(C,0);
        REP(i,0,cycle.size()) 
        {
            bestpath = 0;
            val[i]=DFS(cycle[i]);
            bestval = max(bestval,bestpath);
        }
        lld cursum=0;
        REP(i,0,2*C) {node = cycle[i%C]; ps[i]=(cursum+val[i%C]); cursum+=w[node];}
        Rolling_Max(C-1,2*C);
        REP(i,0,C)
        {
            thisval=rm[i+C-1] -(ps[i]-val[i])+val[i];
            bestval = max(thisval,bestval);
        }
        ans+=bestval;
    }
    cout<<ans<<endl;
    return 0;
}

Compilation message (stderr)

islands.cpp:1: warning: ignoring '#pragma GCC optimization' [-Wunknown-pragmas]
    1 | #pragma GCC optimization ("O3")
      | 
islands.cpp:2: warning: ignoring '#pragma GCC optimization' [-Wunknown-pragmas]
    2 | #pragma GCC optimization ("unroll-loops")
      |

• Subtask 1: 0 / 33
• Subtask 2: 0 / 3
• Subtask 3: 0 / 4
• Subtask 4: 0 / 10
• Subtask 5: 0 / 10
• Subtask 6: 0 / 10
• Subtask 7: 0 / 10
• Subtask 8: 0 / 10
• Subtask 9: 0 / 10