单调栈（模板）

模板题目：poj 3250

Bad Hair Day
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 19813 Accepted: 6788
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 hi (1 ≤ hi ≤ 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 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.

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 cN.
Sample Input

6
10
3
7
4
12
2
Sample Output

5
Source

USACO 2006 November Silver

翻译成人话（中文）就是：
一群高度不完全相同的牛从左到右站成一排，每头牛只能看见它右边的比它矮的牛的发型，若遇到一头高度大于或等于它的牛，则无法继续看到这头牛后面的其他牛。
给出这些牛的高度，要求每头牛可以看到的牛的数量的和并输出

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分析:搞个栈，输入一个数的时候，让他跟栈顶的数比较，比栈顶数大则删掉栈顶，继续比较，直到没比它小的数，压它进栈。这样在压他进栈之前栈中剩下的数就是可以看到它的牛的数量。加起来输出就好。

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#include<iostream>#include<cstdio>#define LL long longusing namespace std;LL num,ans=0,stack[800090],tot=0,n;//数组模拟栈 int main(){cin>>n;for(int i=1;i<=n;i++){cin>>num;while(tot>=1&&stack[tot]<=num){tot--;}ans+=tot;stack[++tot]=num;}cout<<ans;return 0;}