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Bad Hair Day
Time Limit : 4000/2000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)
Total Submission(s) : 48 Accepted Submission(s) : 17
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.
6 10 3 7 4 12 2
5
用a数组倒序创建一个非递增数组（不是递减），当前数与队尾数下标只差既是c[i]
代码：
#include<iostream>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<iomanip>
#include<queue>
#include<cstring>
#include<map>
using namespace std;
typedef long long ll;
#define pi 3.14159265358979
#define inf 0x7fffffff
int n,q[80010];
int a[80010];
void que()
{int i,head=1,tail=0;ll sum=0; q[0]=n+1;for(i=n;i>=1;i--){while(head<=tail&&a[q[tail]]<a[i]) tail--;sum+=q[tail]-i-1;q[++tail]=i;}printf("%I64d\n",sum);
}
int main()
{int i;while(scanf("%d",&n)!=EOF){for(i=1;i<=n;i++)scanf("%d",&a[i]);que();}return 0;
}
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