poj 3250 Bad Hair Day (单调栈）

Bad Hair Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14883		Accepted: 4940

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

单调栈的入门题:单调栈就是维护一个栈，栈中的元素都保持着单调递增或递减的顺序。

题目意思：有n仅仅牛站在一排，给出队伍中每仅仅牛的高度。每仅仅牛仅仅能看到它右边比它矮的牛。求全部的牛能看到的牛数之和。

当我们新增加一个高度值时，假设栈中存在元素小于新增加的高度值。那这个牛肯定看不见这个高度的牛，就把这个元素弹栈。每次增加新元素，并运行完弹出操作后，栈中元素个数便是能够看见这个牛的“牛数”。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
typedef long long ll;
using namespace std;
int main()
{
    int n;
    ll heigh,ans;
    stack<ll>s;
    while(scanf("%d",&n)!=EOF)
    {
        ans=0;
        cin>>heigh;
        s.push(heigh);//入栈
        for(int i=1;i<n;i++)
        {
            cin>>heigh;
            while(!s.empty()&&s.top()<=heigh) //比較栈顶元素和新增加元素的关系
            {
                s.pop();
            }
            ans+=s.size();
            s.push(heigh);
        }
        cout<<ans<<endl;
        while(!s.empty()) s.pop();//空栈
    }
    return 0;
}