POJ 3250 Bad Hair day analog monotonic stack

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) is has a bad hair day! Since Each cow was self-conscious about she messy hairstyle, FJ wants to count the number of other cows so can see the to P of other cows ' heads.

Each cow i have a specified height hi (1≤ hi ≤1,000,000,000) and is standing in a line of cows All facing east (to the right with our diagrams). Therefore, Cow i can see the tops of the heads of cows in front of her (namely cows i+1, i+2, a nd so on), for as long as these cows is strictly shorter than cow I.

Consider this example:

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 are 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 which is the height of cow I.

Output

Line 1: A single integer which is the sum of C1 through CN.

Sample Input

610374122

Sample Output

5

To tell a sequence, to ask for the number of small numbers behind each number and

#include <iostream> #include <stdio.h> #include <string> #include <cstring> #include < algorithm> #include <cmath> #include <queue> #define N 100009typedef 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];//strictly monotonous, do not go to the equal sign            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;}

Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.

POJ 3250 Bad Hair Day simulation monotonic stack