POJ 3250 Bad Hair Day 【单调栈】

Bad Hair Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15952		Accepted: 5387

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
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

恩，题目大意就是说，一头牛 i 只能看到它右边比它矮的牛 j ，但是若在 i 和 j 直间存在比 i 低而比 j 高的牛 k ，则 i 看不到 j 了。总的求的是所有牛能看到的它右边的牛的总和。也就是说，所有牛能被除它之外的牛看到的次数，直接求超时，看人家题解，是单调栈。而单调栈，即是单调增或减的，在当前元素入栈前将栈中比它小（大）或相等的元素删除，如此得到了单调栈。。。。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<stack>
#define maxn 80000
using namespace std;
int rec[maxn];
int main()
{
int n;
unsigned long sum;
while(~scanf("%d",&n))
{
sum=0;
stack<int>s;
int tem;
scanf("%d",&tem);
s.push(tem);
for(int i=1;i<n;++i)
{
scanf("%d",&tem);
while(!s.empty()&&s.top()<=tem)
s.pop();
sum+=s.size();
s.push(tem);
}
printf("%u\n",sum);
}
return 0;
}