POJ 3250 Bad Hair Day(单调栈stack)

Bad Hair Day

Time Limit: 2000MS

Memory Limit: 65536K

Total Submissions: 15958

Accepted: 5392

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 h_i (1 ≤ h_i ≤ 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 c_i denote the number of cows whose hairstyle is visible from cow i; please compute the sum of c₁ through c_N.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 c_N.

Sample Input

610374122

Sample Output

5

题意：一群牛面朝右排列，个高的牛能看见前面个子低的牛的的头发，个子高的牛也能遮住后面个子低的牛的视线。给出这群牛从左到右的身高，求每头牛能看见的发型个数之和。

题解：比赛的时候根本没有想到用单调栈做，暴力果断超时了。这里把问题想成一只牛的发型能被其他几只牛看见？把每只牛压入栈，当下一只将要被压入栈的牛身高比栈顶牛高时，就代表栈顶牛的视线被遮住了，看不见接下来的牛了，则出栈。栈里剩下的牛的个数就是能看见将要入栈牛发型的个数。将每次求得的个数叠加就能得到结果。

代码如下：

#include<cstdio>#include<stack>using namespace std;int main(){int n,i;long long h,sum;//int会错，longlong过了 while(scanf("%d",&n)!=EOF){stack<int>s;//注意更新 sum=0;scanf("%lld",&h);s.push(h);for(i=1;i<n;++i){scanf("%lld",&h);while(!s.empty()&&h>=s.top())//相等也代表看不见前方牛了 s.pop();sum+=s.size();s.push(h);}printf("%lld\n",sum);}return 0;}