POJ-3250 Bad Hair Day 单调栈

Bad Hair Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10300		Accepted: 3458

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 c ₁ through c_N.

Sample Input

6
10
3
7
4
12
2

Sample Output

5

 1 /* 
 2    功能Function Description:     POJ-3250  单调栈
 3    开发环境Environment:          DEV C++ 4.9.9.1
 4    技术特点Technique:
 5    版本Version:
 6    作者Author:                   可笑痴狂
 7    日期Date:                      20120802
 8    备注Notes:
 9        题意：
10            n个牛排成一列向右看，牛i能看到牛j的头顶，当且仅当牛j在牛i的右边并且牛i与牛j之间的所有牛均比牛i矮。
11            设牛i能看到的牛数为Ci，求∑Ci
12 
13        本题正确解法是用栈来做的-----刚开始看的时候表示根本想不到栈
14 
15        单调栈-----所谓单调栈也就是每次加入一个新元素时，把栈中小于等于这个值的元素弹出。
16        接下来回到这道题。求所有牛总共能看到多少牛，可以转化为：这n头牛共能被多少头牛看见。
17        当我们新加入一个高度值时，如果栈中存在元素小于新加入的高度值，那么这些小的牛肯定看不见这个高度的牛(那就看不见这头牛后边的所有牛)，
18        所以就可以把这些元素弹出。每次加入新元素，并执行完弹出操作后，栈中元素个数便是可以看见这个牛的“牛数”~~~。
19 
20        这道题要注意答案可能会超longint，要用int64。
21 */
22 
23 /*代码一：(超时)
24 #include<stdio.h>
25 
26 int main()
27 {
28     int n,i,j,num;
29     int h[80001];
30     __int64 sum;
31     while(~scanf("%d",&n))
32     {
33         sum=0;
34         for(i=0;i<n;++i)
35             scanf("%d",&h[i]);
36         for(i=0;i<n-1;++i)
37         {
38             j=i+1;
39             num=0;
40             while(j<n&&h[i]>h[j++])
41                 ++num;
42             sum+=num;
43         }
44         printf("%lld\n",sum);
45     }
46     return 0;
47 }
48 */
49 
50 //代码二：(单调栈)
51 
52 #include<cstdio>
53 #include<stack>
54 using namespace std;
55 
56 int main()
57 {
58     stack<int> s;
59     __int64 sum;
60     int n,h,t;
61     while(~scanf("%d",&n))
62     {
63         sum=0;
64         scanf("%d",&h);
65         s.push(h);
66         for(int i=1;i<n;++i)
67         {
68             scanf("%d",&t);
69             while(!s.empty()&&t>=s.top())
70                 s.pop();
71             sum+=s.size();
72             s.push(t);          //将当前元素压入栈
73         }
74         printf("%lld\n",sum);
75         while(!s.empty())       //每次用后将栈清空
76             s.pop();
77     }
78     return 0;
79 }