poj-3250 Bad Hair Day(单调栈)

原题链接

Bad Hair Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17707		Accepted: 5971

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

用单调栈求出距离num[i]最近的大于等于num[i]的数num[j], 那么第i头牛可以看到j - i - 1头牛

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <stack>
#define INF 1e9
#define maxn 80005
using namespace std;
typedef long long ll;

int num[maxn];
stack<int> s;
int main(){

// freopen("in.txt", "r", stdin);
int n;

while(scanf("%d", &n) == 1){
while(!s.empty())
s.pop();
for(int i = 0; i < n; i++)
scanf("%d", num+i);
ll ans = 0;
for(int i = n-1; i >= 0; i--){
while(!s.empty() && num[s.top()] < num[i])
s.pop();
if(s.empty())
ans += n - 1 - i;
else
ans += s.top() - i - 1;
s.push(i);
}
printf("%I64d\n", ans);
}
return 0;
}