Bad Hair Day
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 17614 | Accepted: 5937 |
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 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.
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 cN.
Sample Input
6 10 3 7 4 12 2
Sample Output
5
一群高度不完全相同的牛从左到右站成一排，每头牛只能看见它右边的比它矮的牛的发型，若遇到一头高度大于或等于它的牛，则无法继续看到这头牛后面的其他牛。
给出这些牛的高度，要求每头牛可以看到的牛的数量的和。
把要求作一下转换，其实就是要求每头牛被看到的次数之和。这个可以使用单调栈来解决。
从左到右依次读取当前牛的高度，从栈顶开始把高度小于或等于当前牛的高度的那些元素删除，此时栈中剩下的元素的数量就是可以看见当前牛的其他牛的数量。把这个数量加在一起，就可以得到最后的答案了。
单调栈的代码最后竟那么简洁。
1 //2016.8.23
2 #include<cstdio>
3
4 const int N = 80005;
5 int Stack[N], hi, top;
6
7 int main()
8 {
9 int n;
10 long long ans;
11 while(scanf("%d", &n)!=EOF)
12 {
13 ans = top = 0;
14 for(int i = 1; i <= n; i++)
15 {
16 scanf("%d", &hi);
17 while(top>0&&Stack[top-1]<=hi)top--;
18 ans += top;
19 Stack[top++] = hi;
20 }
21 printf("%lld\n", ans);
22 }
23
24 return 0;
25 }