• poj3244(公式题)


    Difference between Triplets
    Time Limit: 3000MS   Memory Limit: 65536K
    Total Submissions: 2476   Accepted: 800

    Description

    For every pair of triplets, T= (IaJaKa) and Tb = (IbJb, Kb), we define the difference value between Ta and Tb as follows:

    D(Ta, Tb) = max {Ia − IbJa − JbKa − Kb} − min {Ia − IbJa − JbKa − Kb}

    Now you are given N triplets, could you write a program to calculate the sum of the difference values between every unordered pair of triplets?

    Input

    The input consists of several test cases. 
    Each test case begins with a line containing an integer N, denotes the number of triplets. Assume that we number the triplets as T1T2, ... , TN. Then, there are following N lines, each line contains three integers, giving the elements of each triplet. 
    A case with N = 0 indicates the end of the input. 

    Output

    For each case, output a line with the sum of difference values between every unordered pair of triplets.

    Sample Input

    2
    1 2 3
    3 2 1
    3
    1 3 2
    4 0 7
    2 2 9
    0
    

    Sample Output

    4
    20
    

    Hint

    Case 1: D(T1,T2)=4 
    Case 2: D(T1,T2)+D(T1,T3)+D(T2,T3)=8+8+4=20 

    You can assume that N, the number of triplets in each case, will not exceed 200,000 and the elements in triplets fit into [-106,106]. 
    The size of the input will not exceed 5 MB. 

    Source


    题目的意思很easy理解,我就不多说了。

    用n^2算法必跪的啦,必需要O(n)

    先要理解max{a,b,c}-min{a,b,c}=(|a-b|+|b-c|+|c-a|)/2

    如果a>b>c,那么max{a-b-c}-min{a-b-c}=a-c

    而(|a-b|+|b-c|+|c-a|)/2=(a-b+b-c+c-a)/2=a-c=max{a-b-c}-min{a-b-c}

    而由位置的对称性,不管a,b,c关系怎样,我们总能够交换其似的a',b',c'保持a'>b'>c',同一时候结果总是一致。

    而| (Ia-ka) -(Ib-kb)| = |(Ia-Ib)-(ka-kb)|

    所以我们对于Ti=(Ia,Ib,Ic)能够先计算Ia-Ib,Ib-Ic,Ic-Ia

    这样我们要计算D(Ti,Tj),则是求(|a-b|+|b-c|+|c-a|)/2.能够保持线性。仅仅要保证a>b, b>c, c>a,则能够将绝对值去掉。

    进行排序,排序后a[i],b[i],c[i],a[i]做头的是i次,被减的是n-i-1次(从0起的坐标)

    这样,ans=sum(i*(a[i]+b[i]+c[i])-(n-i-1)*(a[i]+b[i]+c[i]))

    具体见代码。

    思路的确比較坑。


    #include <iostream>
    #include <algorithm>
    using namespace std;
    #define N 200005
    long long a[N],b[N],c[N];
    int main()
    {
        int i,n;
        long long x,y,z;
        cin.sync_with_stdio(false);
        while(cin>>n,n){
    
                for(i=0;i<n;i++){
                    cin>>x>>y>>z;
        a[i]=x-y,b[i]=y-z,c[i]=z-x;
    
    
                }
                sort(a,a+n);
                sort(b,b+n);
                sort(c,c+n);
                long long ans=0LL;
                for(i=0;i<n;i++){
                    ans+=(i*(a[i]+b[i]+c[i])-(n-i-1)*(a[i]+b[i]+c[i]));
                }
                cout<<ans/2<<endl;
    
        }
        return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/gcczhongduan/p/5153300.html
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