顺手写个单调队列。。。。
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 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 |
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#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
using namespace std;
long long int n,ct[88888];
struct node
{
long long int h,id;
}H[88888];
stack<node> q;
int main()
{
scanf("%I64d",&n);
for(int i=0;i<n;i++) scanf("%I64d",&H[i].h),H[i].id=i;
long long int ans=0;
q.push(H[n-1]);
for(int i=n-2;i>=0;i--)
{
while(!q.empty()&&q.top().h<H[i].h)
{
node D=q.top();
q.pop();
ct[i]+=(1+ct[D.id]);
}
q.push(H[i]);
}
for(int i=0;i<n;i++)
{
ans+=ct[i];
}
printf("%I64d\n",ans);
return 0;
}