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(单调队列) Bad Hair Day -- POJ -- 3250

时间:2015-11-16 10:40:36      阅读:174      评论:0      收藏:0      [点我收藏+]

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http://poj.org/problem?id=3250

 

 

Bad Hair Day
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 15956   Accepted: 5391

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 c1 through cN.

Sample Input

6
10
3
7
4
12
2

Sample Output

5

Source

 

#include<stdio.h>
#define N 80010

int Stack[N];

int main()
{
    int top = 0, i, n, p;
    __int64 sum = 0;

    scanf("%d", &n);

    for(i=1; i<=n; i++)
    {
        scanf("%d", &p);
        while(top>0 && Stack[top]<=p)
            top--;
        sum += top;
        Stack[++top] = p;
    }

    printf("%I64d\n", sum);
    return 0;
}

 

(单调队列) Bad Hair Day -- POJ -- 3250

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原文地址:http://www.cnblogs.com/YY56/p/4968221.html

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