Bad Hair Day
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 23081 | Accepted: 7910 |
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
Source
主要是思维的转化:转化成本牛被几个牛看到。
优先队列。(单调栈也可)
#include <cstdio>
#include <cstring>
#include<queue>
using namespace std;
typedef long long LL;
int main()
{
int n;
LL x;
while(scanf("%d",&n)!=EOF){
priority_queue<int,vector<int>,greater<int> >small;
scanf("%lld",&x);
small.push(x);
LL ans=0;
for(int i=1;i<n;i++){
scanf("%lld",&x);
while(x>=small.top()&&small.empty()==0){
small.pop();
}
ans+=small.size();
small.push(x);
}
printf("%lld\n",ans);
}
}