poj3250 Bad Hair Day

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 c_N.

Sample Input

6
10
3
7
4
12
2

Sample Output

5

一些牛从左到右排列。全部的牛都从左往右看，左边的牛仅仅能看到右边的比它身高严格小的牛的发型。假设被一个大于等于它身高的牛挡住。那么它就不能看到再右边的牛。要求每头牛能够看到其它牛的总数。转化一下，事实上就是求每头牛被看到的总次数。能够用单调栈，每次删除栈中比当前牛的身高小于等于的数。事实上这题也能够看做是单调队列。但由于不用对对首操作。所以可看做退化为了栈。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define maxn 80600
int a[maxn],stack[maxn];
int main()
{
	int n,m,i,j,top;
	__int64 sum;
	while(scanf("%d",&n)!=EOF){
		memset(stack,0,sizeof(stack));
		top=0;sum=0;
		for(i=1;i<=n;i++){
			scanf("%d",&a[i]);
			while(top>0 && a[i]>=stack[top])top--;
			sum+=top;
			stack[++top]=a[i];
		}
		printf("%I64d
",sum);
	}
	return 0;
}