제출 #626151

#제출 시각아이디문제언어결과실행 시간메모리
626151mraronCatfish Farm (IOI22_fish)C++17
3 / 100
1077 ms19680 KiB
#include "fish.h"

#include <vector>
#include <algorithm>
#include <iostream>
using namespace std;

using ll = long long ;

#define sz(x) (int)(x).size()
#define xx first
#define yy second

//dp0[i][j] -> i. oszlopban j-ig elé építve piert max hogyha mindkettő szomszédosnál magasabb
//dp1[i][j] -> i. oszlopban j-ig elé építve piert max hogyha bal oldali számít
//dp2[i][j] -> i. oszlopban j-ig elé építve piert max hogyha jobb oldali számít


long long max_weights(int N, int M, std::vector<int> X, std::vector<int> Y,
                      std::vector<int> W) {
    vector<vector<pair<int,ll>>> lst(N+2, vector<pair<int,ll>>());
    for(int i=0;i<M;++i) {
        lst[X[i]].push_back({Y[i], W[i]});
    }

    for(int i=0;i<=N+1;++i) lst[i].push_back({N+10,0});
    for(auto& i:lst) sort(i.begin(), i.end());
    for(auto& i:lst) 
        for(int j=1;j<sz(i);++j) i[j].yy+=i[j-1].yy;
    
    auto get_sum=[&](int x, int l, int r) -> ll {
        if(l>r) return 0;
        //~ ll sum=0;
        //~ for(int i=0;i<sz(lst[x]);++i) {
            //~ if(l<=lst[x][i].xx && lst[x][i].xx<=r) sum+=lst[x][i].yy;
        //~ }
        //~ cerr<<x<<","<<l<<","<<r<<" -> "<<sum<<"\n";
        auto itL=lower_bound(lst[x].begin(), lst[x].end(), make_pair(l, 0LL));
        auto itR=lower_bound(lst[x].begin(), lst[x].end(), make_pair(r+1, 0LL));
        //~ cerr<<l<<" "<<r<<" -> "<<itL->xx<<" "<<itR->xx<<"\n";
        ll sum=(itR==lst[x].begin()?0:prev(itR)->yy)-(itL==lst[x].begin()?0:prev(itL)->yy);
        return sum;
    };
    
    
    vector<ll> dp0(lst[0].size());
    vector<ll> dp1(lst[0].size());
    vector<ll> dp2(lst[0].size());
    for(int i=1;i<N;++i) {
        vector<ll> ndp0(lst[i].size());
        vector<ll> ndp1(lst[i].size());
        vector<ll> ndp2(lst[i].size());
        
        //dp0
        for(int j=0;j<sz(lst[i]);++j) {
            for(int k=0;k<sz(lst[i-1]);++k) {
                ndp0[j]=max({ndp0[j], dp1[k], dp2[k]+get_sum(i-1, lst[i-1][k].xx, lst[i][j].xx-1)});
            }
        }
        
        //dp1
        for(int j=0;j<sz(lst[i]);++j) {
            for(int k=0;k<sz(lst[i-1]);++k) {
                ndp1[j]=max({ndp1[j], max(dp1[k],dp0[k])+get_sum(i, lst[i][j].xx, lst[i-1][k].xx-1)});
            }
        }
        
        //dp2
        for(int j=0;j<sz(lst[i]);++j) {
            for(int k=0;k<sz(lst[i-1]);++k) {
                ndp2[j]=max({ndp2[j], dp2[k]+get_sum(i-1, lst[i-1][k].xx, lst[i][j].xx-1), dp0[k]});
            }
        }
        //~ cerr<<i<<", 0: "; for(auto j:ndp0) cerr<<j<<" ";cerr<<"\n";
        //~ cerr<<i<<", 1: "; for(auto j:ndp1) cerr<<j<<" ";cerr<<"\n";
        //~ cerr<<i<<", 2: "; for(auto j:ndp2) cerr<<j<<" ";cerr<<"\n\n";
        dp0.swap(ndp0);
        dp1.swap(ndp1);
        dp2.swap(ndp2);
    }
    
    return max({
        *max_element(dp0.begin(), dp0.end()),
        *max_element(dp1.begin(), dp1.end()),
        *max_element(dp2.begin(), dp2.end())
    });
        
}

    
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