POJ 3250 Bad Hair Day（单调栈）

Bad Hair Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 20917		Accepted: 7150

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

Source

USACO 2006 November Silver

 

题目大意：从左往右，每个牛都有一个高度，他能看到右边低于他高度的牛，问总共能看到多少。

 

#include<iostream>
#include<stack>
#include<algorithm>
#include<string>
#include<cstring>
#include<cstdio>
#define ll long long
using namespace std;
ll h[80005];
int r[80005];
stack<ll>s;
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%lld",&h[i]);

    for(int i=n;i>=1;i--)
    {
        while(!s.empty()&&h[s.top()]<h[i]) s.pop();
        if(s.empty()) r[i]=n;
        else r[i]=s.top()-1;
        s.push(i);
    }

    ll ss=0;
    for(int i=1;i<=n;i++)
    {
        ss+=r[i]-i;
        //cout<<i<<" "<<r[i]<<endl;
    }
     printf("%lld
",ss);
    return 0;
}