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  • POJ3250(单调栈)

    Bad Hair Day

    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 17614   Accepted: 5937

    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 hi (1 ≤ h≤ 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

    一群高度不完全相同的牛从左到右站成一排,每头牛只能看见它右边的比它矮的牛的发型,若遇到一头高度大于或等于它的牛,则无法继续看到这头牛后面的其他牛。

    给出这些牛的高度,要求每头牛可以看到的牛的数量的和。

    把要求作一下转换,其实就是要求每头牛被看到的次数之和。这个可以使用单调栈来解决。

    从左到右依次读取当前牛的高度,从栈顶开始把高度小于或等于当前牛的高度的那些元素删除,此时栈中剩下的元素的数量就是可以看见当前牛的其他牛的数量。把这个数量加在一起,就可以得到最后的答案了。

    单调栈的代码最后竟那么简洁。

     1 //2016.8.23
     2 #include<cstdio>
     3 
     4 const int N = 80005;
     5 int Stack[N], hi, top;
     6 
     7 int main()
     8 {
     9     int n;
    10     long long ans;
    11     while(scanf("%d", &n)!=EOF)
    12     {
    13         ans = top = 0;
    14         for(int i = 1; i <= n; i++)
    15         {
    16             scanf("%d", &hi);
    17             while(top>0&&Stack[top-1]<=hi)top--;
    18             ans += top;
    19             Stack[top++] = hi;
    20         }
    21         printf("%lld
    ", ans);
    22     }
    23 
    24     return 0;
    25 }
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  • 原文地址:https://www.cnblogs.com/Penn000/p/5800875.html
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