Bad Hair Day--（单调队列）_综合

Bad Hair Day

Time Limit : 4000/2000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 48 Accepted Submission(s) : 17

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 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, 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 c_i denote the number of cows whose hairstyle is visible from cow i; please compute the sum of c₁ through c_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 c1 through cN.

Sample Input

Sample Output

用a数组倒序创建一个非递增数组（不是递减），当前数与队尾数下标只差既是c[i]

代码：

#include<iostream> #include<string> #include<cstdio> #include<algorithm> #include<cmath> #include<iomanip> #include<queue> #include<cstring> #include<map> using namespace std; typedef long long ll; #define pi 3.14159265358979 #define inf 0x7fffffff int n,q[80010]; int a[80010]; void que() {int i,head=1,tail=0;ll sum=0; q[0]=n+1;for(i=n;i>=1;i--){while(head<=tail&&a[q[tail]]<a[i]) tail--;sum+=q[tail]-i-1;q[++tail]=i;}printf("%I64d\n",sum); } int main() {int i;while(scanf("%d",&n)!=EOF){for(i=1;i<=n;i++)scanf("%d",&a[i]);que();}return 0; }