POJ 3250 Bad Hair Day

Bad Hair Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10202		Accepted: 3412

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

 

 

 

 

大一即将结束了、期末基本考好了、期待暑假的训练，成长

 

本题是栈的应用

把数压入栈，发现有数比栈顶大、就出栈，就是把区间中小的数舍掉，数出栈时计算ci;

思路参考：http://blog.csdn.net/woshi250hua/article/details/7484239

 

#include <iostream>
#include <stdio.h>
#include <string.h>
#define N 80003
using namespace std;
struct node
{
    int index,val;
}re[N];
int main()
{
    int n,i,h;
    __int64 sum;
    node S[N];
    while(scanf("%d",&n)!=EOF)
    {
        h=sum=0;
        scanf("%d",&re[0].val),re[0].index=0;
        S[++h]=re[0];
        for(i=1;i<n;i++)
        {
            scanf("%d",&re[i].val),re[i].index=i;
            while(h>0&&S[h].val<=re[i].val)
            {
               sum+=i-S[h].index-1;
               h--;
            }
            S[++h]=re[i];
        }
       while(h>0&&S[h].val<=2147480000)
            {
               sum+=n-S[h].index-1;
               h--;
            }
        printf("%I64d\n",sum);
    }
    return 0;
}