POJ 3250 Bad Hair Day 【单调栈】

Bad Hair Day

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 15952 Accepted: 5387

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, cowi 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 cowi.

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 cowi; please compute the sum of c₁ throughc_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 cowi.

Output

Line 1: A single integer that is the sum of c₁ throughc_N.

Sample Input

610374122

Sample Output

5

恩，题目大意就是说，一头牛 i 只能看到它右边比它矮的牛 j ，但是若在 i 和 j 直间存在比 i 低而比 j 高的牛 k ，则 i 看不到  j 了。总的求的是所有牛能看到的它右边的牛的总和。也就是说，所有牛能被除它之外的牛看到的次数，直接求超时，看人家题解，是单调栈。而单调栈，即是单调增或减的，在当前元素入栈前将栈中比它小（大）或相等的元素删除，如此得到了单调栈。。。。

#include <iostream>#include<cstdio>#include<cstring>#include<stack>#define maxn 80000using namespace std;int rec[maxn];int main(){    int n;    unsigned long sum;    while(~scanf("%d",&n))    {        sum=0;        stack<int>s;        int tem;        scanf("%d",&tem);        s.push(tem);        for(int i=1;i<n;++i)        {            scanf("%d",&tem);            while(!s.empty()&&s.top()<=tem)                s.pop();            sum+=s.size();            s.push(tem);        }        printf("%u\n",sum);    }    return 0;}