poj--3250--Bad Hair Day(模拟)

Bad Hair Day

Time Limit: 2000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

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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

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Status

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int num[101000];
int main()
{
    int n;
    while(scanf("%d",&n)!= EOF)
    {
        int top = 0;
        long long ans = 0;
        for(int i = 0; i < n; i++)
        {
            int h;
			scanf("%d",&h);
			//保证严格单调递减的序列
            while(top > 0 && num[top] <= h)
                --top;
            ans += top;//记录每一个下标和
            num[++top] = h;
        }
        printf("%lld\n",ans);
    }
    return 0;
}