HDOJ 题目3250 Bad Hair Day(技巧)
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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
610374122
Sample Output
5
Source
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
#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);}}
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