Submission #420459

#TimeUsernameProblemLanguageResultExecution timeMemory
420459MetalPowerCigle (COI21_cigle)C++14
100 / 100
261 ms67296 KiB
#include <bits/stdc++.h> using namespace std; const int MX = 5005; void chmax(int& a, int b){ if(b > a) a = b; } int N, arr[MX], dp[MX][MX]; // Observation #1 : We can easily solve this in O(N^3) // dp(i, j) is the largest beauty of the wall if the highest row includes [i, j] // the transition is we add another row [j + 1, k] for every j + 1 <= k <= N // calculating is easy two pointer // Observation #2 : if the size of row[j + 1, k] is higher than row[i, j] // then for every other k >= current_k // the value of the dp is constant // this is also correct for the opposite // So we can find the positions where the value changes in O(N) for each j // update those positions // and after we update max the current value with the previous value // To make finding range query faster use prefix sum (because l = 0) struct prefix{ int v[MX]; void build(int p){ for(int i = 0; i <= p; i++) v[i] = i == 0 ? dp[i][p] : max(dp[i][p], v[i - 1]); } } ps; int main(){ ios_base::sync_with_stdio(0); cin.tie(NULL); cout.tie(NULL); cin >> N; for(int i = 0; i < N; i++) cin >> arr[i]; for(int p = 0; p < N; p++){ int i = p, k = p + 1, cnt = -1, bot = 0, top = 0; ps.build(p); while(i >= 0 && k < N){ if(bot == top){ cnt++; chmax(dp[p + 1][k], ps.v[i] + cnt); bot += arr[i--]; top += arr[k++]; }else if(bot < top){ bot += arr[i--]; }else if(top < bot){ top += arr[k++]; } } for(int j = p + 1; j < N; j++) chmax(dp[p + 1][j], dp[p + 1][j - 1]); } int ans = 0; for(int p = 0; p < N; p++) chmax(ans, dp[p][N - 1]); cout << ans << '\n'; }
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