标签:get 复杂 时间 规划 lse template c++ min .com
[题目链接]
https://www.lydsy.com/JudgeOnline/problem.php?id=3831
[算法]
单调队列优化动态规划
时间复杂度 : O(N)
[代码]
#include<bits/stdc++.h> using namespace std; #define MAXN 1000010 const int inf = 2e9; int n; int a[MAXN] , f[MAXN]; template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); } template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); } template <typename T> inline void read(T &x) { T f = 1; x = 0; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -f; for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - ‘0‘; x *= f; } inline int solve(int k) { deque< int > q; for (int i = 1; i <= n; i++) f[i] = inf; q.clear(); for (int i = 1; i <= n; i++) { while (!q.empty() && i - q.front() > k) q.pop_front(); if (q.empty()) f[i] = 0; else f[i] = f[q.front()] + (a[q.front()] <= a[i]); while (!q.empty() && ((f[i] < f[q.back()]) || ((f[i] == f[q.back()]) && a[i] >= a[q.back()]))) q.pop_back(); q.push_back(i); } return f[n]; } int main() { read(n); for (int i = 1; i <= n; i++) read(a[i]); int T; read(T); while (T--) { int x; read(x); printf("%d\n" , solve(x)); } return 0; }
标签:get 复杂 时间 规划 lse template c++ min .com
原文地址:https://www.cnblogs.com/evenbao/p/9911902.html