DP/單調隊列：：poj3250 Bad Hair Day_poj3250time limit exceeded-优快云博客

本文探讨了一种高效的方法来解决复杂问题，通过引入单调队列技术优化算法性能，详细解释了其工作原理和应用实例，展示了在特定场景下显著提高计算效率的可能性。

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题目：

Some of Farmer John's N cows (1 ≤ N ≤ 80,000) are having a bad hair day! Since each cow is self-conscious about her messy hairstyle, FJ wants to count the number of other cows that can see the top of other cows' heads.

Each cow i has a specified height h_i (1 ≤ h_i≤ 1,000,000,000) and is standing in a line of cows all facing east (to the right in our diagrams). Therefore, cowi can see the tops of the heads of cows in front of her (namely cows i+1, i+2, and so on), for as long as these cows are strictly shorter than cowi.

Let c_i denote the number of cows whose hairstyle is visible from cowi; please compute the sum ofc₁ through c_N.

分析：用DP去做，r[i]表示从右数第一个比a[i]大的数的前一个数的位置，则当a[i]<a[r[i]+1]时有r[i]=r[r[i]+1]. 求出了数组r[i]之后累加即可。一开始我把遍历的顺序弄错了，为for i=1..N，结果“Time Limit Exceeded”因为这里的DP根本不起作用，还是重复计算了！改了之后就对了，代码如下（Memory: 876K，Time: 641MS）

#include <iostream>
using namespace std;
unsigned long a[80001];
unsigned long r[80001];
unsigned long sum=0;

int main()
{
	int N;
	cin>>N;
	for(int i=1;i<=N;i++){
		cin>>a[i];
		r[i]=i;
	}
	for(int i=N-1;i>=0;i--){
		while(r[i]<N && a[r[i]+1]<a[i]){
			r[i]=r[r[i]+1];
		}
		sum+=r[i]-i;
	}

	cout<<sum;

	return 0;
}

本题最好的方法还是用单调队列，因为上面DP的实现效率不是很高，也不容易分析。单调队列的话，其实和在hist中找最大矩形的方法一样的。这里不在重复。