poj 3250 Bad Hair Day（单调栈）

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本文介绍了一道名为“BadHairDay”的编程题解，题目要求计算每头牛能看到前方多少头比它矮的牛。通过使用单调栈的数据结构，从右到左遍历数组并记录比当前元素小的数量，最终得出所有可见数量之和。

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题目链接：http://poj.org/problem?id=3250点击打开链接

Bad Hair Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 19144		Accepted: 6515

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 c ₁ through c_N.

Sample Input

Sample Output

从右往左遍历每个位置记录下比他小的有几个用单调栈可以实现

#include <iostream>
#include <stdio.h>
#include <limits.h>
#include <stack>
#include <algorithm>
using namespace std;
long long int a[100000];
long long int b[100000];
stack<long long int > s;
int main()
{
    long long int n=0;
    scanf("%lld",&n);
    for(long long int i=0;i<n;i++)
    {
        scanf("%lld",&a[i]);
    }
    for(long long int i=n-1;i>=0;i--)
    {
        int ans=0;
        while(1)
        {
            if(s.empty())
            {
                s.push(i);
                b[i]=ans;
                break;
            }
            else
            {
               if(a[s.top()]<a[i])
               {
                   ans+=b[s.top()];
                   ans++;
                   s.pop();
               }
               else
               {
                   s.push(i);
                   b[i]=ans;
                   break;
               }
            }
        }
    }
    long long int sum=0;
    for(long long int i=0;i<n;i++)
        {
            sum+=b[i];
            //cout << b[i] << endl;
        }
    cout << sum;
}