제출 #755014

#제출 시각아이디문제언어결과실행 시간메모리
755014Red_InsideCatfish Farm (IOI22_fish)C++17
0 / 100
929 ms165964 KiB
#include "fish.h" #include <bits/stdc++.h> //#pragma GCC optimize("O1,O2,O3,Ofast,unroll-loops") #define ll long long #define f first #define s second #define pb push_back #define mp make_pair #define o cout<<"BUG"<<endl; #define FOR(i, j, n) for(int j = i; j < n; ++j) #define forn(i, j, n) for(int j = i; j <= n; ++j) #define nfor(i, j, n) for(int j = n; j >= i; --j) #define all(v) v.begin(), v.end() //#define ld long double #define ull unsigned long long using namespace std; const int maxn=3e5+10,LOG=17,mod=1e9+7; int block = 650, timer = 0; const double eps = 1e-9; //mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); #define IOS ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0); #define bt(i) (1 << (i)) //#define int ll const long double inf=2e18; #define y1 yy #define pii pair <int, int> int n, m, x[maxn], y[maxn], w[maxn]; ll a[3010][3010], pref[3010][3010], dp2[3010][3010]; ll dp[3010][3010], Dp[3010], ans; long long max_weights(int N, int M, vector <int> X, vector <int> Y, vector <int> W) { n = N; m = M; forn(1, i, m) { x[i] = X[i-1]+1; y[i] = Y[i-1]+1; w[i] = W[i-1]; a[x[i]][y[i]] = w[i]; } forn(1, i, n) { forn(1, j, n) { pref[i][j] = pref[i][j - 1] + a[i][j]; } } forn(1, i, n) { dp[0][i] = -inf; } forn(1, i, n) { ll tek = 0; if(i >= 2) { forn(1, j, n) { tek = max(tek, dp[i-1][j] - pref[i-1][j]); dp[i][j] = max(dp[i][j], pref[i-1][j] + tek); } } // forn(1, j, n) // cout << dp[i][j] << " "; // cout << endl; if(i >= 2) { tek = 0; forn(1, j, n) { tek = max(tek, dp[i-2][j]); dp[i][j] = max(dp[i][j], pref[i-1][j] + tek); } } if(i >= 3) { tek = 0; nfor(1, j, n) { dp[i][j] = max(dp[i][j], tek - pref[i-1][j]); tek = max(tek, dp[i-2][j] + pref[i-1][j]); } } if(i >= 3) { forn(1, j, n) { dp[i][j] = max(Dp[i-3] + pref[i-1][j], dp[i][j]); } } tek = -inf; nfor(1, j, n) { dp2[i][j] = max(dp2[i][j], tek - pref[i][j]); // cout << i << " " << dp2[i][j] << " " << tek << endl; tek = max(tek, max(dp2[i-1][j], dp[i-1][j]) + pref[i][j]); Dp[i] = max(Dp[i], dp2[i][j] + pref[i+1][j]); } forn(1, j, n) { Dp[i] = max(Dp[i], dp[i][j] + pref[i+1][j]); } ans = max(ans, Dp[i]); // cout << Dp[i] << " " << i << endl; } return ans; }
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