Bad Hair Day （单调队列）

Bad Hair Day （单调队列）

Problem 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 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.

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 c₁ through c_N.

Sample Input

610374122

Sample Output

5

题目大概：

题目是羊的身高不一样，羊从左向右站着，只能看到在自己前面并且比自己低的羊。问每只羊看到的羊的加和是多少。

思路：

这道题和例题的求矩形面积有相似点，当然这个题只是求矩形面积的小小的一部分内容，只是求一点向一个方向的连续的点的数量并求和，在题量上，和求矩形面积没法比。这个题只需要一个单调队列，不断维护，并顺便求出和便好了。

代码：

#include #include using namespace std;int n;int a[100000];int r[100000];int q[100000];long long su; int ri=0;int go(){ for(int i=1;i<=n;i++) { while(ri!=0&&a[i]>=r[ri])ri--; su+=ri; r[++ri]=a[i]; }}int main(){ scanf("%d",&n); for(int i=1;i<=n;i++) { scanf("%d",&a[i]); } go(); cout<

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