POJ 3250 Bad Hair Day 模拟单调栈

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 hi (1 ≤ h≤ 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  c 1 through  cN.

Sample Input

6
10
3
7
4
12
2

Sample Output

5


告诉一个序列,求每个数后面比他小的个数和

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#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(tt<n&&a[i]>a[tt+1]) tt=b[tt+1];//严格单调,不去等号
            b[i]=tt;
        }

//        for(int i=1;i<=n;i++)
//        cout<<b[i]<<" ";

        for(int i=1;i<=n;i++)
        {
            ans+=b[i]-i;
        }
        printf("%I64d\n",ans);

    }

    return 0;
}






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