POJ 3250 Bad Hair Day

Bad Hair Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17727		Accepted: 5981

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

 

    所以的牛都是面朝右边的，牛 i 能看到牛 j 当且仅当 i<j，h[ i ] > h[k], k=i , i+1, .. ,j。

    维护一个单调栈，从左往右扫描h[i]，(1) 当栈为空时，把h[i]压入栈中；(2)当栈非空时，不断弹出栈顶元素，知道当前h[i]小于栈顶元素，把h[i]压入栈。 这样，h[i]被压入栈前栈的大小，就是能看到第 i 头牛的其它牛数量，累加即可。

 

#include <iostream>
#include <algorithm>
#include <map>
#include <vector>
#include <functional>
#include <string>
#include <cstring>
#include <queue>
#include <stack>
#include <set>
#include <cmath>
#include <cstdio>
using namespace std;
#define IOS ios_base::sync_with_stdio(false)
typedef long long LL;
const int INF = 0x3f3f3f3f;
const double PI=4.0*atan(1.0);

const int maxn=80000;
LL ans;
int n,h;
int main()
{
    while(scanf("%d",&n)!=EOF){
        stack<int> s;
        ans=0;
        for(int i=0;i<n;i++){
            scanf("%d",&h);
            while(!s.empty()&&h>=s.top()) s.pop();
            ans+=s.size();
            s.push(h);
        }
        printf("%lld\n",ans);
    }
}