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 ≤ hi ≤ 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
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i.
Output
Sample Input
6 10 3 7 4 12 2
Sample Output
5
一些牛从左到右排列。全部的牛都从左往右看，左边的牛仅仅能看到右边的比它身高严格小的牛的发型。假设被一个大于等于它身高的牛挡住。那么它就不能看到再右边的牛。要求每头牛能够看到其它牛的总数。转化一下，事实上就是求每头牛被看到的总次数。能够用单调栈，每次删除栈中比当前牛的身高小于等于的数。事实上这题也能够看做是单调队列。但由于不用对对首操作。所以可看做退化为了栈。
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define maxn 80600
int a[maxn],stack[maxn];
int main()
{
int n,m,i,j,top;
__int64 sum;
while(scanf("%d",&n)!=EOF){
memset(stack,0,sizeof(stack));
top=0;sum=0;
for(i=1;i<=n;i++){
scanf("%d",&a[i]);
while(top>0 && a[i]>=stack[top])top--;
sum+=top;
stack[++top]=a[i];
}
printf("%I64d\n",sum);
}
return 0;
}
