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| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 17614 | Accepted: 5937 |
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.
6 10 3 7 4 12 2
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 }
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原文地址：http://www.cnblogs.com/Penn000/p/5800875.html
