POJ 3250 Bad Hair Day 模擬單調棧

 

Bad Hair Day Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 14989   Accepted: 4977

 

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

 

告訴一個序列，求每個數後面比他小的個數和

 

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define N 100009
typedef long long ll;

using namespace std;

ll a[N];

int main()
{
    int n;
    while(~scanf(%d,&n))
    {
        for(int i=1;i<=n;i++)
            scanf(%I64d,&a[i]);

        ll ans=0;
        int b[N];
        b[n]=n;
        for(int i =n-1;i>=1;i--)
        {
            int tt=i;
            while(tta[tt+1]) tt=b[tt+1];//嚴格單調，不去等號
            b[i]=tt;
        }

//        for(int i=1;i<=n;i++)
//        cout<