Submission #994312

#	Time	Username	Problem	Language	Result	Execution time	Memory
994312		thinknoexit	Catfish Farm (IOI22_fish)	C++17	3 / 100	98 ms	36176 KiB

fish

#include "fish.h"
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
vector<pair<int, int>> pos[100100];
struct DP {
    ll in, de;
    DP() {
        in = de = 0;
    }
} dp[1001000];
int w[1001000];
int n, m;
ll max_weights(int N, int M, vector<int> X, vector<int> Y, vector<int> W) {
    n = N, m = M;
    for (int i = 0;i < m;i++) {
        w[i + 1] = W[i];
        pos[X[i] + 1].push_back({ Y[i] + 1, i + 1 });
    }
    int idx = m, p;
    ll mx, sum;
    for (int i = 2;i <= n;i++) {
        sort(pos[i].begin(), pos[i].end());
        pos[i].push_back({ 0, ++idx });
        pos[i].push_back({ n + 1, ++idx });
        sort(pos[i].begin(), pos[i].end());
        // in -> in
        {
            p = 0;
            mx = 0;
            for (int j = 0;j < (int)pos[i].size();j++) {
                while (p < (int)pos[i - 1].size() && pos[i - 1][p].first <= pos[i][j].first - 1) {
                    int prev = pos[i - 1][p++].second;
                    mx = max(mx, dp[prev].in) + w[prev];
                }
                dp[pos[i][j].second].in = mx;
            }
        }
        // de -> de , in -> de
        {
            p = pos[i - 1].size() - 1;
            mx = 0;
            for (int j = pos[i].size() - 1;j >= 0;j--) {
                while (p >= 0 && pos[i - 1][p].first >= pos[i][j].first) {
                    int prev = pos[i - 1][p--].second;
                    mx = max({ mx, dp[prev].in, dp[prev].de });
                }
                mx += w[pos[i][j].second];
                dp[pos[i][j].second].de = mx;
            }
        }
        // in -> de, de -> in
        if (i >= 3) {
            // h[i-2] <= h[i]
            {
                int mid = 0;
                p = 0;
                mx = 0, sum = 0;
                for (int j = 0;j < (int)pos[i].size();j++) {
                    while (p < (int)pos[i - 2].size() && pos[i - 2][p].first <= pos[i][j].first) {
                        int prev = pos[i - 2][p++].second;
                        mx = max({ mx ,dp[prev].in, dp[prev].de });
                    }
                    while (mid < (int)pos[i - 1].size() && pos[i - 1][mid].first <= pos[i][j].first - 1) {
                        sum += w[pos[i - 1][mid++].second];
                    }
                    dp[pos[i][j].second].in = max(dp[pos[i][j].second].in, mx + sum);
                }
            }
            // h[i-2] >= h[i]
            {
                int mid = pos[i - 1].size() - 1;
                mx = 0, sum = 0;
                for (int j = 0;j <= mid;j++) sum += w[pos[i - 1][j].second];
                p = pos[i - 2].size() - 1;
                for (int j = pos[i].size() - 1;j >= 0;j--) {
                    while (p >= 0 && pos[i - 2][p].first >= pos[i][j].first) {
                        int prev = pos[i - 2][p].second;
                        int now = pos[i - 2][p].first - 1;
                        while (mid >= 0 && pos[i - 1][mid].first > now) {
                            sum -= w[pos[i - 1][mid--].second];
                        }
                        mx = max({ mx, dp[prev].in + sum, dp[prev].de + sum });
                        p--;
                    }
                    dp[pos[i][j].second].in = max(dp[pos[i][j].second].in, mx);
                }
            }
        }
    }
    ll ans = 0;
    for (int i = 1;i <= idx;i++) {
        ans = max(ans, dp[i].in);
        ans = max(ans, dp[i].de);
    }
    return ans;
}

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