Bad Hair Day_单调栈

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

 

【题意】给出一些牛的高度，站成一排，往右看，求所有牛能看见的牛的数量之和。

【思路】单调栈。从左到右取当前牛的高度，从栈顶开始把高度小于或等于当前牛的高度的那些元素删除，此时栈中剩下的元素的数量就是可以看见当前牛的其他牛的数量。把这个数量加在一起，即答案。

#include<iostream>
#include<stdio.h>
#include<stack>
#include<string.h>
using namespace std;
int main()
{
    stack<int>st;
    long long int sum;
    int n,h;
    while(~scanf("%d",&n))
    {
        while(!st.empty())
        {
            st.pop();
        }
        sum=0;
        scanf("%d",&h);
        st.push(h);
        for(int i=1;i<n;i++)
        {
            scanf("%d",&h);
            while(!st.empty()&&h>=st.top())
                st.pop();
            sum+=st.size();
            st.push(h);
        }
        printf("%lld\n",sum);

    }
    return 0;
}