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
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i.
Output
Sample Input
6 10 3 7 4 12 2
Sample Output
5
【题意】给出一些牛的高度,站成一排,往右看,求所有牛能看见的牛的数量之和。
【思路】单调栈。从左到右取当前牛的高度,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,即答案。
#include<iostream> #include<stdio.h> #include<stack> #include<string.h> using namespace std; int main() { stack<int>st; long long int sum; int n,h; while(~scanf("%d",&n)) { while(!st.empty()) { st.pop(); } sum=0; scanf("%d",&h); st.push(h); for(int i=1;i<n;i++) { scanf("%d",&h); while(!st.empty()&&h>=st.top()) st.pop(); sum+=st.size(); st.push(h); } printf("%lld ",sum); } return 0; }