POJ 3250 Bad Hair Day -- 栈-优快云博客

本文链接：https://blog.youkuaiyun.com/spark_007/article/details/8929952

本文介绍如何使用单调队列解决一个涉及牛只高度排序的问题，计算每只牛可以看见其他牛头的数量之和。

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Bad Hair Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 11340		Accepted: 3832

Description

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.

Consider this example:

        =
=       =
=   -   =         Cows facing right -->
=   =   =
= - = = =
= = = = = =
1 2 3 4 5 6

Cow#1 can see the hairstyle of cows #2, 3, 4
Cow#2 can see no cow's hairstyle
Cow#3 can see the hairstyle of cow #4
Cow#4 can see no cow's hairstyle
Cow#5 can see the hairstyle of cow 6
Cow#6 can see no cows at all!

Let c_i denote the number of cows whose hairstyle is visible from cowi; please compute the sum ofc₁ throughc_N.For this example, the desired is answer 3 + 0 + 1 + 0 + 1 + 0 = 5.

Input

Line 1: The number of cows, N.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i.

Output

Line 1: A single integer that is the sum of c ₁ through c_N.

Sample Input

Sample Output

Source

USACO 2006 November Silver

构造一个单调队列，也可以用栈实现，更加简单。

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
long long int stack[80005],t;
int main()
{
	int n,i,top=-1;
	unsigned long ans = 0;
	while(scanf("%d",&n)!=EOF && n!=0)
	{
		memset(stack,'0',sizeof(stack));
		top = -1;ans = 0;
	for(i=0;i<n;i++)
	{
		scanf("%lld",&t);
		if(top == -1)stack[++top]=t;
		else
		{
			while(top >= 0 && stack[top]<=t)top--;
			stack[++top] = t;
			ans += top;
		}
	}
		printf("%lu\n",ans);
	}
     return 0;
}