Submission #637731

#	Time	Username	Problem	Language	Result	Execution time	Memory
637731		someone	Catfish Farm (IOI22_fish)	C++17	18 / 100	748 ms	189344 KiB

fish

#include "fish.h"
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
 
struct Pier {
    int x, y;
    ll fishL = 0, fishAct = 0, fishR = 0, val[2] = {0, 0};
};
 
struct Fish {
    bool req;
    int y;
    ll pds;
};
 
const ll N = 1e5 + 42, INF = 1e15 + 42;
 
ll sz[N];
vector<Fish> fish[N];
vector<Pier> pier[N];
unordered_map<ll, ll> nb[N];
 
ll max_weights(int n, int M, vector<int> X, vector<int> Y, vector<int> W) {
    for(int i = 0; i < M; i++) {
        Y[i] += 1;
        fish[X[i]].push_back({false, Y[i], W[i]});
        for(int j = max(X[i]-2, 0); j < min(X[i]+3, n); j++)
            fish[j].push_back({true, Y[i], 0});
        pier[X[i]].push_back({X[i], Y[i]});
        if(X[i] != 0)
            pier[X[i]-1].push_back({X[i], Y[i]});
        if(X[i] != n-1)
            pier[X[i]+1].push_back({X[i], Y[i]});
    }
 
    for(int i = 0; i < n; i++) {
        sort(fish[i].begin(), fish[i].end(),
        [](Fish a, Fish b) {
            if(a.y == b.y) {
                if(a.req == b.req)
                    return false;
                return b.req;
            }
            return a.y < b.y;
        });
        ll sum = 0;
        for(Fish f : fish[i]) {
            if(f.req) {
                nb[i][f.y] = sum;
            } else {
                sum += f.pds;
            }
        }
    }
    for(int i = 0; i < n; i++) {
        for(Pier& p : pier[i]) {
            if(i != 0)
                p.fishL = nb[i-1][p.y];
            p.fishAct = nb[i][p.y];
            if(i != n-1)
                p.fishR = nb[i+1][p.y];
        }
    }
 
    for(int i = 0; i < n; i++)
        sort(pier[i].begin(), pier[i].end(),
        [](Pier a, Pier b) {
            return a.y < b.y;
        });
    for(int i = 0; i < n; i++)
        sz[i] = (ll)pier[i].size();
    
    ll maxi = 0;
    for(int i = 0; i < n; i++) {
        for(Pier& p : pier[i]) {
            p.val[0] = maxi + p.fishL - p.fishAct;
            p.val[1] = maxi + p.fishL + p.fishR;
        }
 
        ll id[2] = {0, 0}, maxiLL = -INF, maxiL = -INF;
        for(Pier& p : pier[i]) {
            if(i > 0) {
                while(id[0] < sz[i-1] && pier[i-1][id[0]].y <= p.y) {
                    maxiL = max(maxiL, pier[i-1][id[0]].val[0]);
                    id[0]++;
                }
            }
            if(i > 1) {
                while(id[1] < sz[i-2] && pier[i-2][id[1]].y <= p.y) {
                    maxiLL = max(maxiLL, pier[i-2][id[1]].val[1] - pier[i-2][id[1]].fishR - pier[i-2][id[1]].fishAct);
                    id[1]++;
                }
            }
            p.val[0] = max(p.val[0], max(maxiL, maxiLL) - p.fishAct + p.fishL);
            p.val[1] = max(p.val[1], max(maxiL, maxiLL) + p.fishL + p.fishR);
        }
        if(i == 0) {
            id[0] = id[1] = -1;
        } else if(i == 1) {
            id[0] = sz[i-1]-1;
            id[1] = -1;
        } else {
            id[0] = sz[i-1]-1;
            id[1] = sz[i-2]-1;
        }
        maxiLL = maxiL = -INF;
        reverse(pier[i].begin(), pier[i].end());
 
        for(Pier& p : pier[i]) {
            if(i > 0) {
                while(id[0] >= 0 && pier[i-1][id[0]].y >= p.y) {
                    maxiL = max(maxiL, pier[i-1][id[0]].val[1]);
                    id[0]--;
                }
            }
            if(i > 1) {
                while(id[1] >= 0 && pier[i-2][id[1]].y >= p.y) {
                    maxiLL = max(maxiLL, pier[i-2][id[1]].val[1]);
                    id[1]--;
                }
            }
            p.val[0] = max(p.val[0], maxiLL - p.fishAct);
            p.val[1] = max(p.val[1], max(maxiLL, maxiL - p.fishAct) + p.fishR);
        }
 
        reverse(pier[i].begin(), pier[i].end());
 
        if(i > 1) {
            for(Pier p : pier[i-2])
                maxi = max(maxi, max(p.val[0], p.val[1]));
        }
    }
    for(Pier p : pier[n-1])
        maxi = max(maxi, max(p.val[0], p.val[1]));
    for(Pier p : pier[n-2])
        maxi = max(maxi, max(p.val[0], p.val[1]));
    
    return maxi;
}

• Subtask 1: 3 / 3
• Subtask 2: 6 / 6
• Subtask 3: 9 / 9
• Subtask 4: 0 / 14
• Subtask 5: 0 / 21
• Subtask 6: 0 / 17
• Subtask 7: 0 / 14
• Subtask 8: 0 / 16