NYOJ练习题 how many hairstyles can they see?

时间限制：1000 ms  |  内存限制：65535 KB

描述

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). 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.

 

For example:

 

There are six cows, heights are 10  2  7  3  12  2.

 

Now 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!

 

输入

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.

输出

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

样例输入

6

10  2  7  3  12  2

样例输出

5

题意：有n头牛面向右排队，每头牛只能看到右边比它矮的牛，右边比它高的牛会挡住视线；如果比它高的牛的右边还有比它矮的牛，它同样看不到那些牛。求这n头牛能看到的其他牛的数量的总和。

解法：使用栈。如果栈顶的数大于当前元素，说明栈顶的牛可以看见当前的牛，则把当前的元素压入栈中；否则，说明栈顶这头牛看不到当前位置之后的牛了，就可以计算出栈顶的牛可以看到的牛的个数，并让栈顶元素出栈，然后继续比较栈顶元素，直到栈顶元素大于当前位置，把当前元素压入栈中。

 

#include<cstdio>  
#include<cstring>  
#include<algorithm>  
using namespace std;  
const int N = 8e4 + 10;  
int sta[N], pos[N];  
int main()  
{  
    int n, i, a;  
    while(~scanf("%d",&n))  
    {  
        int top = 0;  
        scanf("%d",&a);  
        sta[++top] = a;  
        pos[top] = 1;  
        int ans = 0;  
        for(i = 2; i <= n; i++)  
        {  
            scanf("%d",&a);  
            while(top > 0 && sta[top] <= a)  
            {  
                ans += i - pos[top] - 1;  //求出的ans是top位置的牛能看到的牛的数量  
                top--;  
            }  
            sta[++top] = a;  
            pos[top] = i;  
        }  
        top--; //最后一头牛一头也看不到  
        while(top > 0)  
        {  
            ans += n - pos[top];  
            top--;  
        }  
        printf("%d\n",ans);  
    }  
    return 0;  
}  

 

 

/*当找到一个比栈顶元素大的数时，就让栈顶元素出栈，直到栈空，把当前元素压入栈中*/  

 

补充：软件开发 , C++ ,