1686: Bad Hair Day（单调栈）

1686: Bad Hair Day 

时间限制(普通/Java):2000MS/20000MS     内存限制:65536KByte
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 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.

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 c_N.

样例输入

 

6
10
3
7
4
12
2

样例输出

 5

题目来源

USACO November 2006

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 }

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