제출 #639377

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
639377		SlavicG	Catfish Farm (IOI22_fish)	C++17	0 / 100	762 ms	2097152 KiB

fish

#include "fish.h"
#include "bits/stdc++.h"
using namespace std;
using ll = long long;

const ll inf = 1e16;
long long max_weights(int n, int m, vector<int> x, vector<int> y, vector<int> w) {
    vector<vector<ll>> p(n, vector<ll>(n, 0));
    for(int i = 0; i < m; ++i) {
        p[x[i]][y[i]] += w[i];
    }
    for(int i = 0; i < n; ++i) {
        for(int j = 1; j < n; ++j) {
            p[i][j] += p[i][j - 1];
        }
    }

    vector<vector<vector<ll>>> dp(n + 1, vector<vector<ll>>(n + 1, vector<ll>(2, -inf)));
    for(int i = 0; i <= n; ++i) {
        dp[0][i][0] = dp[0][i][1] = 0;
    }

    //0 - increasing, 1 - decreasing
    for(int i = 1; i < n; ++i) {
        vector<ll> max0(n + 1, -inf), max1(n + 1, -inf);
        for(int lastlen = 0; lastlen <= n; ++lastlen) {
            dp[i][0][0] = max({dp[i][0][0], dp[i - 1][lastlen][0], dp[i - 1][lastlen][1]});
            dp[i][0][1] = max({dp[i][0][1], dp[i - 1][lastlen][0], dp[i - 1][lastlen][1]});
            if(lastlen) {
                dp[i][0][1] = max(dp[i][0][1], dp[i - 1][lastlen][1] + p[i][lastlen - 1]);
                dp[i][0][1] = max(dp[i][0][1], dp[i - 1][lastlen][0] + p[i][lastlen - 1]);
            }
            if(lastlen) max0[lastlen] = max0[lastlen - 1];
            if(lastlen) max1[lastlen] = max1[lastlen - 1];
            max0[lastlen] = max(max0[lastlen], dp[i - 1][lastlen][0] - (lastlen ? p[i - 1][lastlen - 1] : 0));
            max1[lastlen] = max(max1[lastlen], dp[i - 1][lastlen][0] + (lastlen ? p[i][lastlen - 1] : 0));
            max1[lastlen] = max(max1[lastlen], dp[i - 1][lastlen][1] + (lastlen ? p[i][lastlen - 1] : 0));
            for(int curlen = 1; curlen <= n; ++curlen) {
                if(curlen >= lastlen) {
                    dp[i][curlen][0] = max(dp[i][curlen][0], p[i - 1][curlen - 1] + max0[lastlen]);
                }
                if(curlen <= lastlen) {
                    dp[i][curlen][1] = max(dp[i][curlen][1], max1[curlen] - p[i][curlen - 1]);
                }
            }
        }
    }
    ll ans = 0;
    for(int i = 0; i <= n; ++i) {
        ans = max({ans, dp[n - 1][i][0], dp[n - 1][i][1]});
    }
    return ans;
}
/*
5 4
0 2 5
1 1 2
4 4 1
3 3 3
*/

• Subtask 1: 0 / 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