| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 14883 | Accepted: 4940 |
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
Output
Sample Input
6 10 3 7 4 12 2
Sample Output
5
单调栈的入门题:单调栈就是维护一个栈，栈中的元素都保持着单调递增或递减的顺序。
题目意思：有n只牛站在一排，给出队伍中每只牛的高度，每只牛只能看到它右边比它矮的牛，求所有的牛能看到的牛数之和。
当我们新加入一个高度值时，如果栈中存在元素小于新加入的高度值，那这个牛肯定看不见这个高度的牛，就把这个元素弹栈。每次加入新元素，并执行完弹出操作后，栈中元素个数便是可以看见这个牛的“牛数”。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
typedef long long ll;
using namespace std;
int main()
{
int n;
ll heigh,ans;
stack<ll>s;
while(scanf("%d",&n)!=EOF)
{
ans=0;
cin>>heigh;
s.push(heigh);//入栈
for(int i=1;i<n;i++)
{
cin>>heigh;
while(!s.empty()&&s.top()<=heigh) //比较栈顶元素和新加入元素的关系
{
s.pop();
}
ans+=s.size();
s.push(heigh);
}
cout<<ans<<endl;
while(!s.empty()) s.pop();//清空栈
}
return 0;
}
原文地址：http://blog.csdn.net/whjkm/article/details/46352257
