SGU 485 Arrays

485. Arrays

Time limit per test: 1.75 second(s)
Memory limit: 262144 kilobytes

input: standard
output: standard

You are given a sequence of 3· N integers (X₁, X₂, ·s, X_3· N). Create three sequences (A₁, A₂, ·s, A_N), (B₁, B₂, ·s, B_N) and (C₁, C₂, ·s, C_N) such that:

• each of the integers from 1 to 3· N belongs to exactly one of the sequences A, B or C;
• the value of  is the largest possible.

Input

Constraints on N	Constraints on T
1 ≤ N ≤ 10	1 ≤ T ≤ 1000
11 ≤ N ≤ 15	1 ≤ T ≤ 100
16 ≤ N ≤ 20	1 ≤ T ≤ 10
21 ≤ N ≤ 25	T = 1

The input file contains T test cases, all having the same value of N. The first line of the input file contains the integers T and N, constrained as shown in the adjacent table. Each of the following T lines describes one test case and contains 3· N integers, the members of the sequence X. All these values are in the range from 0 to 1000.

Output

The output file should consist of T lines. Each line should contain the largest possible value of S for the corresponding test case from the input file.

Example(s)

sample input	sample output
1 2 4 1 8 2 0 5	46

Note. The maximal value is attained by taking A = (1, 3), B = (2, 5), C = (4, 6).

 

 题意：给出一组数，将这些数分成三组，记为A,B,C，求出满足sigma[(Ai-Bi)*Ci]的最大值。

sl: 首先考虑下B ,很显然B中的元素应该是最小的N个，在考虑A,C很容易看出A,C应该是满足

A>=C的关系。

所以我们只需要枚举A,C就好了。

其中包含2个优化

优化1：在计算的过程中应满足全局平均值最小。

优化2：不等式 （ai-bi）*ci+(aj-bj)*cj - {(ai-bi)*aj+(ci-bj)*cj   } >0成立 即满足：

（ci-aj）*(ai-bi-cj)>0 成立。

所以ci>aj 时 ai-bi>cj

ci<aj时 ai-bi<cj

详见代码。

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 using namespace std;
 5 typedef long long LL;
 6 const int maxn = 76;
 7 int T, N;
 8 int ans, a[maxn], va[maxn], vc[maxn];
 9 bool vis[maxn];
10 bool test(int cur) {
11     for (int i=1; i<cur; ++i) {
12         if (abs(va[cur])>abs(vc[i]) && abs(vc[cur])<abs(va[i])-abs(a[3*N-i+1])) {
13             return false;
14         }
15         if (abs(va[cur])<abs(vc[i]) && abs(vc[cur])>abs(va[i])-abs(a[3*N-i+1])) {
16             return false;
17         }
18     }
19     return true;
20 }
21 void dfs(int cur, int last, int val) {
22     if (cur==N+1) {
23         ans = max(ans, val);
24         return;
25     }
26     for (int i=cur; i<=2*N; ++i) {
27         if (!vis[i]) {
28             vis[i] = 1;
29             va[cur] = a[i];
30             for (int j=max(last+1, i+1); j<=N*2; ++j) {
31                 if (!vis[j]) {
32                     vc[cur] = a[j];
33                     vis[j] = 1;
34                     int netVal = val+(a[i]-a[3*N-cur+1])*a[j];
35                     if (netVal*N>ans*cur) {
36                         if (test(cur)) {
37                             dfs(cur+1, j, netVal);
38                         }
39                     }
40                     vis[j] = 0;
41                 }
42             }
43             vis[i] = 0;
44             break;
45         }
46     }
47 }
48 int main()
49 {
50     scanf ("%d%d", &T, &N);
51     while (T--) {
52         for (int i=1; i<=3*N; ++i) {
53             scanf ("%d", &a[i]);
54             a[i] = -a[i];
55         }
56         sort(a+1, a+3*N+1);
57         ans = 0;
58         dfs(1, 0, 0);
59         printf ("%d\n", ans);
60     }
61     return 0;

62 }