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 6Cow#1 can see the hairstyle of cows #2, 3, 4Cow#2 can see no cow's hairstyleCow#3 can see the hairstyle of cow #4Cow#4 can see no cow's hairstyleCow#5 can see the hairstyle of cow 6Cow#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 c 1 through cN.Sample Input
6 10 3 7 4 12 2Sample Output
5Source
USACO 2006 November Silver
题目大意：从左往右，每个牛都有一个高度，他能看到右边低于他高度的牛，问总共能看到多少，，
反着想，看这个牛能被多少牛看到，要是左边的有比他低或等于他就不被看到，直接去掉就好，
ac代码
[cpp] view plain copy print? #include<stdio.h> #include<stack> #include<string> #include<iostream> using namespace std; int main() { int n; while(scanf("%d",&n)!=EOF) { int a[100010]; int i; __int64 ans=0; stack<int>s; for(i=0;i<n;i++) { scanf("%d",&a[i]); } for(i=0;i<n;i++) { while(!s.empty()&&s.top()<=a[i]) s.pop(); ans+=s.size(); s.push(a[i]); } printf("%I64d\n",ans); } }
转载于:https://www.cnblogs.com/yzm10/p/7197507.html
