不会打数学公式，只能扣图了……

题目翻译

题目描述

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 hi (1 ≤ hi ≤ 1,000,000,000) and is standing in a line of cows all facing east (to the right in our diagrams). Therefore, cow i 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 cow i.

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 ci denote the number of cows whose hairstyle is visible from cow i; please compute the sum of c1 through cN.For this example, the desired is answer 3 + 0 + 1 + 0 + 1 + 0 = 5.

原题链接

Solve

1.单调队列。

2.既然是找比自己高度小的，就用单调递减。

3.每找到一个比自己高度小的，c[]加1，并加上其的c[];

Tip

1.单调递减，所以不会找到比自己高度小的奶牛的高度更小的奶牛(向队首遍历)，从而解释了Slove中的步骤3.

2.不开long long 见祖宗。

Code

#include<iostream>
#include<cstdio>
#define N 80010
#define ll long long
using namespace std;  //标准开头
ll n,ans;  //n-奶牛数，ans-答案
ll q[N][3],h[N],c[N];  //q-队列 q[][1]表示奶牛的高度 q[][2]表示奶牛的序号，h-高度，c-同题中的c[]
ll head=1,tail=0;  //头指针，尾指针
int main()
{
	scanf("%lld",&n);
	for(ll i=1;i<=n;i++) scanf("%lld",&h[i]);
	for(ll i=n;i>=1;i--){
		while(head<=tail&&h[i]>q[tail][1]){
			c[i]++;
			c[i]+=c[q[tail][2]];
			tail--;
		}
		tail++;
		q[tail][1]=h[i];
		q[tail][2]=i;
		ans+=c[i];
	}
	printf("%lld",ans);  //完美输出
	return 0;
}