poj3250

 

Bad Hair Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4588		Accepted: 1385

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

610374122

Sample Output

5

     这个题目如用两个for()循环，是肯定会超时的，这道题我是用

栈的思想做出来的（惭愧，不是自己想出来），感觉蛮不错的，所以

贴出来，供以后察看，总结。

代码如下：

#include<stdio.h>
#include<iostream>
using namespace std;

int stack[80001];
int main()
{
    int i,n,temp,top=-1;
    unsigned ans=0;
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
         scanf("%d",&temp);
         if(top==-1)  stack[++top]=temp;
         else
         {
             while(temp>=stack[top]&&top>=0)  top--;
             stack[++top]=temp;
             ans+=top;
         }             
    }
    printf("%u/n",ans);
    system("pause");
    return 0;
}