USACO Bad Hair Day
USACO Bad Hair Day
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
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 c
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
題解:
單調棧模板。
程式碼:
#include<cstdio>
#include<stack>
#define int long long
using namespace std;
const int maxn=80002;
int n,ans;
stack<int> st;
int h[maxn];
signed main()
{
scanf("%lld",&n);
for(int i=1;i<=n;i++)
scanf("%lld",&h[i]);
h[n+1]=1e9;
for(int i=1;i<=n+1;i++)
{
while((!st.empty())&&(h[i]>=h[st.top()]))
{
ans=ans+i-st.top()-1;
st.pop();
}
st.push(i);
}
printf("%lld\n",ans);
return 0;
}