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_a= (I_a, J_a, K_a) and T_b= (I_b, J_b, K_b), we define the difference value between T_a and T_bas follows:

D(T_a, T_b) = max {I_a − I_b, J_a − J_b, K_a − K_b} − min {I_a − I_b, J_a − J_b, K_a − K_b}

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 T₁, T₂, ... , T_N. 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

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

POJ Monthly--2007.07.08, Yuan, Xinhao

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;
}