Bad Hair Day POJ - 3250 （单调栈）

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
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 cN.
Sample Input
6
10
3
7
4
12
2
Sample Output
5

题意：

题意大概就是几头牛排排站，统一面朝右（如下图），但是每头牛的身高不同，设Ci为奶牛i可见发型的奶牛数，请计算C1到Cn的总和。也就是求奶牛i能比右边多少个奶牛高，但有遮挡作用，即加入奶牛1身高为7 奶牛2身高为8 奶牛3身高为2 ，那么奶牛2挡住了奶牛1，奶牛1则看不到奶牛3。

思路：

这个题考察的是单调栈的知识。

关于栈的简单知识点：
stack<int> sta;
将x入栈：sta.push(x);
出栈：sta.pop();
判断栈的大小： sta.size();
判断栈是否为空：sta.empty();

样例的结果计算：
第一次：先将10入栈
第二次：3与10比较，10大于3，不需要pop出10，此时栈内元素只有一个（10），再将3入栈
第三次：7与栈顶元素3比较，3小于7，pop出3，10与7比较，10大于7，不需pop ，此时栈内元素只有一个（10），再将7入栈
第四次：4与7比较，4小于7，不需pop，此时栈内元素有两个（10，7），再将4入栈
第五次：12与4比较，4小于12，pop出4，同理，pop出7，10，此时栈内为空，元素为0个，再将12入栈
第六次：2与12比较，2小于12，不需pop，栈内有1个元素（12），2入栈
所以最终结果为：1+1+2+0+1=5

大概思路
按照顺序依次入栈，当栈内的元素小于等于即将入栈的元素，则pop掉，计算当前栈内元素的数量，然后再将即将入栈的元素入栈，最后将每一次计算的栈内元素的数量加起来即为所求结果

代码：

#include<stdio.h>
#include<stack>
#include<string.h>
#include<algorithm>
using namespace std;
stack<long long int>q;
int main()
{
    int n;
    scanf("%d",&n);
    long long m,s=0;
    scanf("%lld",&m);
    q.push(m);
    for(int i=1; i<n; i++)
    {
        scanf("%lld",&m);
        while(q.size()&&q.top()<=m)//当栈不为空且栈顶元素小于等于m
            q.pop();
        s+=q.size();//删除比m小的数后计算栈的大小然后再将m入栈
        q.push(m);
    }
    printf("%lld\n",s);
    return 0;
}