Luogu

單調隊列板子題。

很明顯\[Ans=\max_{1≤i≤n} \{\sum_{j=1}^i P_i-\min_{i-m≤k<i} \{\sum_{j=1}^k P_j\} \}\]

單調隊列維護即可。

#include <iostream> #include <cstdio> #include <queue> const int max_n = 500000 + 5; const int inf = 0x7f7f7f7f;int N, M, Ans = -inf; int sum[max_n];std::deque <int> q;inline int read() {register int x = 0, v = 1;register char ch = getchar();while(!isdigit(ch)) {if(ch == '-') v = -1;ch = getchar();}while(isdigit(ch)){x = (x << 1) + (x << 3) + ch - '0';ch = getchar();}return x * v; } int main() {N = read();M = read();for(int i = 1; i <= N; ++i) sum[i] = sum[i - 1] + read();for(int i = 1; i <= N; ++i){while(!q.empty() && sum[q.back()] > sum[i]) q.pop_back();q.push_back(i);while(!q.empty() && i - q.front() > M) q.pop_front();Ans = std::max(Ans, sum[i] - sum[q.front()]); }printf("%d\n", Ans);return 0; }

轉載于:https://www.cnblogs.com/zcdhj/p/8457905.html

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