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题意：#
For every pair of triplets, Ta = (Ia, Ja, Ka) and T__b = (Ib, Jb, Kb), we define the difference value between Ta and_T__b_ as follows:
D(Ta,_ Tb_) = max {Ia − Ib, Ja − Jb, Ka − Kb} − min {Ia − Ib, Ja − Jb, Ka − 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?
思路：转化要求的式子，如果把_Ia_ − Ib, Ja − Jb, Ka − Kb 看成数轴上的点，所求的式子就变成了求三个点构成的线段的距离。#
**设X=Ia − Ib,Y=Ja − Jb,Z=Ka − Kb，那么该距离D = （|X-Y| + |Y-Z| + |Z-X| ）/2（该式子是此题最关键的一部，可以通过画图直观得到） **#
D=|(ia-ja)-(ib-jb)| + |(ja-ka)-(jb-kb)| + | (ka-za)+(kb-zb) |#
设a = ia-ja,b =ja-ka,c = ka-ia，然后分别排序类似于bzoj1604_拆点求曼哈顿距离#
考虑第i个a，对于其他的n-1个a，有i-1个比它小，n-i个比它大，因此对答案的贡献为(i-1)个a[i]和 (n-i)个-a[i]#
b,c同理。#
代码实现
1/* ***********************************************
2Author :111qqz
3Created Time :2016年11月17日 星期四 20时05分54秒
4File Name :code/hdu/3244.cpp
5************************************************ */
6#include <cstdio>
7#include <cstring>
8#include <iostream>
9#include <algorithm>
10#include <vector>
11#include <queue>
12#include <set>
13#include <map>
14#include <string>
15#include <cmath>
16#include <cstdlib>
17#include <ctime>
18#define fst first
19#define sec second
20#define lson l,m,rt<<1
21#define rson m+1,r,rt<<1|1
22#define ms(a,x) memset(a,x,sizeof(a))
23typedef long long LL;
24#define pi pair < int ,int >
25#define MP make_pair
26using namespace std;
27const double eps = 1E-8;
28const int dx4[4]={1,0,0,-1};
29const int dy4[4]={0,-1,1,0};
30const int inf = 0x3f3f3f3f;
31const int N=2E5+7;
32int n;
33int x,y,z;
34int a[N],b[N],c[N];
35int main()
36{
37 #ifndef ONLINE_JUDGE
38 freopen("code/in.txt","r",stdin);
39 #endif
40 while (~scanf("%d",&n))
41 {
42 if (n==0) break;
43 for ( int i = 1; i <= n ; i++)
44 {
45 scanf("%d%d%d",&x,&y,&z);
46 a[i] = x - y;
47 b[i] = y - z;
48 c[i] = z - x;
49 }
50 sort(a+1,a+n+1);
51 sort(b+1,b+n+1);
52 sort(c+1,c+n+1);
53 LL ans = 0 ;
54 for ( int i = 1 ; i <= n ; i++)
55 {
56 ans += 1LL*(i-1)*a[i];
57 ans +=-1LL*(n-i)*a[i];
58 ans += 1LL*(i-1)*b[i];
59 ans +=-1LL*(n-i)*b[i];
60 ans += 1LL*(i-1)*c[i];
61 ans +=-1LL*(n-i)*c[i];
62 }
63 printf("%lld\n",ans/2);
64 }
65 #ifndef ONLINE_JUDGE
66 fclose(stdin);
67 #endif
68 return 0;
69}