POJ 3250 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 hermessy hairstyle, FJ wants to count the number of other cows that can see thetop 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 cowsall facing east (to the right in our diagrams). Therefore, cow i can seethe 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 cowswhose hairstyle is visible from cow i; please compute the sum of c1through 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 ofcow i.
Output
Line 1: A single integer thatis the sum of c1 through cN.
Sample Input
6
10
3
7
4
12
2
Sample Output
5
题目简介：一群牛排队，头向右。问每头牛能看到前面多少头牛的头（即前方连续有多少头牛比自己矮）。然后算出其总数。
方法：双端队列。队列中存储牛的数组下标，将高度按降序排列。当处理的数据比队列末尾下标对应牛的高度大时，在总数上加上队尾计数器的值，用count保存下此时队尾计数器的值，然后删去队尾。同时在删去后的队尾的计数器上加上（count+1）。数据处理完后，从队尾开始进行类似操作。
例如：队列为10，3，此时处理7，3<7，3不能看到任意一头牛即<3,0>，sum+=0，将3删除，同时10的计数器上加1，表示10能看到1头牛（<10,1>），10>7。继续处理后面的数据，此时队列为10，7。处理4，4小于7，继续处理。处理12，12>4，4不能看到任意一头牛，即<4,0>，sum+=0，将4删除，同时在7的计数器上加1，即<7,1>。继续比较，7<12，sum+=1，
10的计数器上加上2,即<10,3>。10<12，sum+=3。10之前没有数据，删去10即可。接下来处理2，2<12，不进行任何操作。然后数据处理完毕。接着从队尾开始，删去2，<2.,0>,sum+=0。然后在12上加1即<12,1>,sum+=1，删去12。队列为空，完毕。
语言能力有限。。。。。。
#include<iostream>
#include<deque>
using namespace std;
struct node
{
long long int high;
int count;
}cow[80001];
int main()
{
int n, i;
scanf("%d",&n);
for(i = 0;i<n;i++)
{
scanf("%lld",&cow[i].high);
}
deque<int> q;
q.clear();
long long sum = 0;
for(i = 0;i<n;i++)
{
int count = 0;
while(!q.empty()&&cow[i].high>=cow[q.back()].high)
{
cow[q.back()].count += count;
sum += cow[q.back()].count;
count = cow[q.back()].count + 1;
q.pop_back();
}
if(!q.empty())
{
cow[q.back()].count += count;
}
q.push_back(i);
}
int count = 0;
while(!q.empty())
{
cow[q.back()].count += count;
sum += cow[q.back()].count;
count = cow[q.back()].count + 1;
q.pop_back();
}
printf("%lld\n",sum);
system("pause");
return 0;
}