Submission #1042164

#	Time	Username	Problem	Language	Result	Execution time	Memory
1042164		VMaksimoski008	Catfish Farm (IOI22_fish)	C++17	52 / 100	315 ms	507124 KiB

fish

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

ll pref[3005][3005], dp[3005][3005][2], pmx1[3005][3005], smx1[3005][3005], pmx2[3005][3005], smx2[3005][3005];
 
ll max_weights(int N, int M, vector<int> X, vector<int> Y, vector<int> W) {
    for(int i=0; i<M; i++) pref[X[i]+1][Y[i]+1] += W[i];
    for(int i=1; i<=N; i++)
        for(int j=1; j<=N; j++) pref[i][j] += pref[i][j-1];
    for(int i=1; i<=N; i++)
        for(int j=0; j<=N; j++) pmx1[i][j] = smx1[i][j] = pmx2[i][j] = smx2[i][j] = -1e18;

    pmx1[1][0] = dp[1][0][1] - pref[1][0];
    smx1[1][N] = max(dp[1][N][0], dp[1][N][1]) + pref[2][N];
    for(int i=1; i<=N; i++) pmx1[1][i] = max(pmx1[1][i-1], dp[1][i][1] - pref[1][i]);
    for(int i=N-1; i>=0; i--) smx1[1][i] = max(smx1[1][i+1], max(dp[1][i][0], dp[1][i][1]) + pref[2][i]);
    for(int i=1; i<=N; i++) pmx2[1][i] = max(pmx2[1][i-1], max(dp[1][i][0], dp[1][i][1]));
    for(int i=N-1; i>=0; i--) smx2[1][i] = max(smx2[i][i+1], max(dp[1][i][0], dp[1][i][1]));

    for(int i=2; i<=N; i++) {
        for(int j=0; j<=N; j++) {
            dp[i][j][0] = max(dp[i][j][0], smx1[i-1][j] - pref[i][j]);
            dp[i][j][1] = max({ dp[i][j][1], pmx1[i-1][j] + pref[i-1][j] });
            if(i >= 3) dp[i][j][0] = max({ dp[i][j][0], pmx2[i-2][j] + pref[i-1][j], smx2[i-2][j] + pref[i-1][j] });
            if(i >= 3) dp[i][j][1] = max({ dp[i][j][1], pmx2[i-2][j] + pref[i-1][j], smx2[i-2][j] + pref[i-1][j] });

            if(j) pmx1[i][j] = pmx1[i][j-1];
            if(j) pmx2[i][j] = pmx2[i][j-1];
            pmx1[i][j] = max(pmx1[i][j], dp[i][j][1] - pref[i][j]);
            pmx2[i][j] = max(pmx2[i][j], max(dp[i][j][0], dp[i][j][1]));
        }

        smx1[i][N] = max(dp[i][N][0], dp[i][N][1]) + pref[i+1][N];
        for(int j=N-1; j>=0; j--) smx1[i][j] = max(smx1[i][j+1], max(dp[i][j][0], dp[i][j][1]) + pref[i+1][j]);
        smx2[i][N] = max(dp[i][N][0], dp[i][N][1]);
        for(int j=N-1; j>=0; j--) smx2[i][j] = max(smx2[i][j+1], max(dp[i][j][0], dp[i][j][1]));
    }

    ll ans = 0;
    for(int i=1; i<=N; i++)
        for(int j=1; j<=N; j++) ans = max({ ans, dp[i][j][0] + pref[i+1][j], dp[i][j][1] + pref[i+1][j] });
    return ans;
}

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