• POJ 3250 Bad Hair Day(单调栈)


    Bad Hair Day
    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 20917   Accepted: 7150

    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

    Source

     
    题目大意:从左往右,每个牛都有一个高度,他能看到右边低于他高度的牛,问总共能看到多少。
     
    #include<iostream>
    #include<stack>
    #include<algorithm>
    #include<string>
    #include<cstring>
    #include<cstdio>
    #define ll long long
    using namespace std;
    ll h[80005];
    int r[80005];
    stack<ll>s;
    int main()
    {
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%lld",&h[i]);
    
        for(int i=n;i>=1;i--)
        {
            while(!s.empty()&&h[s.top()]<h[i]) s.pop();
            if(s.empty()) r[i]=n;
            else r[i]=s.top()-1;
            s.push(i);
        }
    
        ll ss=0;
        for(int i=1;i<=n;i++)
        {
            ss+=r[i]-i;
            //cout<<i<<" "<<r[i]<<endl;
        }
         printf("%lld
    ",ss);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/caiyishuai/p/13271058.html
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