题目描述

Some of Farmer John's N cows () 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 hi () 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 ci denote the number of cows whose hairstyle is visible from cow i; please compute the sum of c1 through cN.For this example, the desired is answer 3 + 0 + 1 + 0 + 1 + 0 = 5.

输入描述:

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.

输出描述:

Line 1: A single integer that is the sum of c1 through cN.

示例1

输入
6
10
3
7
4
12
2
输出
5

解答

题目描述

一群身高不等的奶牛排成一排，向右看，每个奶牛只能看到身高小于自己的奶牛发型，问这些奶牛能够看到的发型总和是多少
思路：利用单调栈（栈中元素从栈顶往下越来越大）的思想，可以计算出，每只奶牛能被他前面的多少只奶牛看到自己的发型，就反向得到了奶牛看到的发型总和
借用别人的解释：首先弹出元素是因为它右边相邻牛比它高（看不到它的头发并且挡住了该牛的视线），“挡住”这个关键字很重要，这就是说该牛已经看不到后面的其他牛的头发啦，而思路一是按顺序比较身高，统计每头牛前面能看到它头发的牛数，既然该牛望不到后面，那么把它出栈对整个结果没有影响。
例如：10 3 7 4 12 2
①Height_List：10 入栈首身高
②3<10，不弹出，num=0+1（当前栈中元素数），3入栈后 Height_List：10 3
③7 > 3（3 弹出） 7<10（10保留），栈中剩 10 ，num=1+1，7入栈后 Height_List：10 7
④4 < 7，没有弹出，栈中10 7都能看到4，num=2+2 ，4入栈后 Height_List：10 7 4
⑤12大于栈中全部元素，表示栈中所有元素看不到12，全部出栈，num不变，12入栈后 Height_List：12
⑥2 < 12，没有弹出，12能看到2，num=4+1
最后结果num=5
代码如下
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <stack>

using namespace std;

stack<long long> st;
int a[80005];

int main()
{
    int i,n;
    long long sum = 0;
    scanf("%d",&n);
    for(i = 0;i < n;i++)
        scanf("%d",&a[i]);
    for(i = 0;i < n;i++)
    {
        while(!st.empty() && st.top() <= a[i])	//如果栈顶元素小于即将入栈元素，那么就不断出栈
            st.pop();
        sum += st.size();//st.size()表示将要入栈的元素，它后面能看到他发型的奶牛个数
        st.push(a[i]);
    }
    printf("%lld\n",sum);
    return 0;
}

来源：键盘上Dance