https://atcoder.jp/contests/abc285/tasks/abc285_e
Let the first day be weekend, then day
Let
that is in each stage we have two decisions:
- assign day
weekday, then find the max productivity and update day - assign day
weekend, then find the max productivity and update interval
// compile: g++ -o data data.cpp -O3 -std=gnu++20 -Wall -Wextra -Wshadow -D_GLIBCXX_ASSERTIONS -ggdb3 -fmax-errors=2 -DLOCAL // run: ./data < data.in #include <bits/stdc++.h> using namespace std; #pragma GCC optimize("O3,unroll-loops") #pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt") #ifdef LOCAL #include <debug/codeforces.h> #define debug(x...) {_variables(#x);_print(x);} #else #define debug(x...) #endif template<typename...Args> void print_(Args...args){((cout<<args<<" "),...)<<endl;} #define rep(i,a,b) for(int i=(a);i<(b);++i) #define sz(v) ((int)(v).size()) #define print(...) print_(__VA_ARGS__); #define INTINF (int)(9223372036854775807) #define int long long #define MAXN 5010 int n, a[MAXN], s[MAXN]; int dp[MAXN][MAXN]; int get(int l, int r) { int days = r - l + 1; return 2 * s[days >> 1] + ((days & 1) * (a[(days >> 1) + 1])); } int32_t main() { ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); cin >> n; for (int i = 1; i <= n; ++i) cin >> a[i], s[i] = s[i-1] + a[i]; for (int i = 2; i <= n; ++i) { for (int j = 1; j < i; ++j) { dp[i][i] = max(dp[i][i], dp[j][j] + get(j+1, i-1)); dp[i][j] = dp[i-1][j] + a[min(i-j, n+1-i)]; } } int ans = 0; for (int i = 1; i <= n; ++i) ans = max(ans, dp[n][i]); cout << ans << endl; return 0; }
The problem is easy but took me more than 20 mins.
- When encountering cyclic DP problems, try to break the cycle.
- Think carefully about the states, whether there are multiple ways to express it? try to use the one with the simplest state transition function.