POJ 3250 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 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

题意：问你每个人向右看能看到多少人（i<j&&i,j之间的高度均小于i，j）
分析：如果正者来，对于每个点向右找符合条件的显然比价复杂，我们可以反着来，对于
每一个点，向左找能和这个点搭配的=-=，显然维护一个单调递减栈即可。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<set>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
const int INF=0x3f3f3f3f;
typedef long long LL;
const int maxn=1e6+100;
int s[maxn],h[maxn];
int n;
int main()
{
    while(~scanf("%d",&n))
    {
        REPF(i,1,n)
          scanf("%d",&h[i]);
        int top=0;
        LL sum=0;
        for(int i=1;i<=n;i++)
        {
            while(top>0&&s[top]<=h[i])
                --top;
            sum+=top;
            s[++top]=h[i];
        }
        printf("%I64d\n",sum);
    }
}