1686: Bad Hair Day 
总提交: 68 测试通过:31
描述
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.
输入
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.
输出
Line 1: A single integer that is the sum of c1 through cN.
样例输入
6
10
3
7
4
12
2
样例输出
5
题目来源
解题思路： 题目意思 第i个能看到知道比其大或者相等之前的所有牛！ 维护一个单调递减的栈,如果新来的元素比栈顶元素小那栈中的元素都可以看到它 相反那之前的哪些都没用了 都被这个高的给挡住了！
1 // 2 #include <iostream> 3 #include <cstring> 4 #include <algorithm> 5 #include <cstdio> 6 #include <vector> 7 using namespace std; 8 9 typedef long long ll; 10 const int N=80005; 11 int sta[N]; 12 int n,d; 13 ll res; 14 15 int main(){ 16 ios::sync_with_stdio(false); 17 int top=1; 18 cin>>n; 19 cin>>d; 20 sta[1]=d; 21 for(int i=2;i<=n;i++){ 22 cin>>d; 23 while(d>=sta[top]&&top) top--; 24 res+=top; //在其前面的都能看到他 25 sta[++top]=d; 26 } 27 cout << res << endl; 28 return 0; 29 }