POJ 3250 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 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
Source
#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;
}