poj 3250 Bad Hair Day

Bad Hair Day

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 15968 Accepted: 5395

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 h_i (1 ≤ h_i ≤ 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 c_i denote the number of cows whose hairstyle is visible from cow i; please compute the sum of c₁ through c_N.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 c₁ through c_N.

Sample Input

610374122

Sample Output

5

刚开始一直在用循环优化，几乎无奈了，最后万万没想到竟然用的是栈，哎，没想到。。。。

思路： 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。       接下来回到这道题。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”~~~。

        还要注意long long ..

  贴下代码：

    
#include<stdio.h>#include<iostream>#include<algorithm>#include<stack>using namespace std;int main(){int n;while(scanf("%d",&n)!=EOF) { int i,k; __int64 sum; sum=0; stack<int > s; scanf("%d",&k); s.push(k); for(i=1;i<n;i++) { scanf("%d",&k); while(!s.empty() &&k>=s.top()) s.pop(); sum+=s.size();            s.push(k); } printf("%I64d\n",sum); while(!s.empty()) s.pop(); }}