Bad Hair Day
/*
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 */
n个牛排成一列向右看,牛i能看到牛j的头顶,当且仅当牛j在牛i的右边并且牛i与牛j之间的所有牛均比牛i矮。设牛i能看到的牛数为Ci,求∑Ci
单调栈-----所谓单调栈也就是每次加入一个新元素时,把栈中小于等于这个值的元素弹出。
求所有牛总共能看到多少牛,可以转化为:这n头牛共能被多少头牛看见。
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,那么这些小的牛肯定看不见这个高度的牛(那就看不见这头牛后边的所有牛),所以就可以把这些元素弹出。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”。
这道题要注意答案可能会超longint,要用int64。
#include <iostream>
#include <stack>
#include <stdio.h>
#include <algorithm>
using namespace std; int main()
{
int n;
while(cin >> n)
{
stack<int> s;
int be_cow;
cin >> be_cow;
s.push(be_cow);
long long ans = ;
for(int i = ; i < n; i++)
{
int cow;
cin >> cow;
while(!s.empty() && cow >= s.top())
s.pop();
ans += s.size();
s.push(cow);
}
cout << ans << endl;
}
return ;
}
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