Submission #825044

#	Time	Username	Problem	Language	Result	Execution time	Memory
825044		PixelCat	Catfish Farm (IOI22_fish)	C++17	35 / 100	64 ms	9996 KiB

fish

#include "fish.h"

#ifdef NYAOWO
#include "grader.cpp"
#endif

#include <bits/stdc++.h>
#define For(i, a, b) for(int i = a; i <= b; i++)
#define Forr(i, a, b) for(int i = a; i >= b; i--)
#define F first
#define S second 
#define all(x) x.begin(), x.end()
#define sz(x) ((int)x.size())
#define eb emplace_back
#define int LL
using namespace std;
using i32 = int32_t;
using LL = long long;
using pii = pair<int, int>;

inline void chmax(int &x, int val) { x = max(x, val); }

const int MAXN = 300;
const int INF = 1e15;

int n, m;
int pre[MAXN + 10][MAXN + 10];
int dp1[MAXN + 10][MAXN + 10];
int dp2[MAXN + 10][MAXN + 10];

int sum(int x, int u, int d) {
    if(u < d) return 0;
    int res = pre[x][u];
    if(d) res -= pre[x][d - 1];
    return res;
}

long long max_weights(i32 N, i32 M, vector<i32> X, vector<i32> Y,
                      vector<i32> W) {
    n = N;
    m = M;

    memset(pre, 0, sizeof(pre));
    For(i, 0, m - 1) {
        pre[X[i] + 1][Y[i] + 1] += W[i];
    }
    For(i, 1, n) For(j, 1, n) {
        pre[i][j] += pre[i][j - 1];
    }

    For(i, 0, n + 1) For(j, 0, n + 1) {
        dp1[i][j] = dp2[i][j] = -INF;
    }
    dp2[0][0] = 0;

    For(i, 1, n + 1) {
        int mx = -INF;
        For(j, 1, n) {
            dp1[i][j] = dp2[i - 1][0] + sum(i - 1, j, 0);
            // chmax(dp1[i][j], mx + pre[i - 1][j]);
            // chmax(dp1[i][j], dp1[i - 1][j]);
            // chmax(dp1[i][j], dp2[i - 1][j]);
            // chmax(mx, dp1[i - 1][j] - pre[i - 1][j + 1]);
            // chmax(mx, dp2[i - 1][j] - pre[i - 1][j + 1]);
            For(j2, 1, j) chmax(dp1[i][j], dp1[i - 1][j2] + sum(i - 1, j, j2 + 1));
            For(j2, 1, j) chmax(dp1[i][j], dp2[i - 1][j2] + sum(i - 1, j, j2 + 1));
        }
        mx = -INF;
        Forr(j, n, 0) {
            chmax(mx, dp2[i - 1][j]);
            dp2[i][j] = dp1[i - 1][j] + sum(i, j, 0);
            chmax(dp2[i][j], mx + pre[i][j] - pre[i - 1][j]);
            // For(j2, j, n) chmax(dp2[i][j], dp2[i - 1][j2] + sum(i, j, 0) - sum(i - 1, j, 0));
        }
    }

    // Forr(j, n, 0) For(i, 1, n + 1) {
    //     cerr << dp1[i][j] << " \n"[i == n + 1];
    // }
    // cerr << "\n";
    // Forr(j, n, 0) For(i, 1, n + 1) {
    //     cerr << dp2[i][j] << " \n"[i == n + 1];
    // }

    int ans = 0;
    For(j, 0, n) chmax(ans, dp2[n + 1][j]);
    return ans;
}

/*

5 4
0 2 5
1 1 2
4 4 1
3 3 3

8

*/

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