POJ 3250 Bad Hair Day 算法学习：单调栈

Bad Hair Day

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 18534 Accepted: 6266

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 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 cowi.

Output

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

Sample Input

610374122

Sample Output

5

题意：

有N只牛向右看，只能看到第一只比他高的牛之前的牛。问每头牛能看到的牛的总数是多少？

方法：单调栈

所谓单调栈，就是在每次押入一个元素之前，将栈中所有大于（小于）他的元素弹出栈，这样就保证栈有一定的大小顺序。

每次读入一只牛的高度，就将栈中比他矮的牛弹出，然后将这只牛压入栈。因为比他矮的牛将看不到他及其右边的牛。所以，此时栈的大小便是能看到当前牛的牛的只数。这时再将这只牛的身高压入栈中。

代码：

//By Sean Chen#include <iostream>#include <cstdio>#include <cstring>using namespace std;int Stack[800005],top;        //单调栈和栈顶指针int main(){    int n,hi;    long long cnt=0;      //答案需要用long long 表示    scanf("%d",&n);    for (int i=0;i<n;i++)    {        scanf("%d",&hi);        while (top>0 && hi>=Stack[top])        //单调栈操作，题目要求高度大于才能看到，所以弹栈的条件应是hi>=Stack[top]            top--;        cnt+=top;        Stack[++top]=hi;    }    printf("%lld\n",cnt);    return 0;}