Hdu 3480 Division (slope optimization)

DivisionTime Limit: 10000/5000 MS (Java/Others) Memory Limit: 999999/400000 K (Java/Others)Total Submission (s): 2676 Accepted Submission (s): 1056

Problem DescriptionLittle D is really interested in the theorem of sets recently. There's a problem that confused him a long time.
Let T be a set of integers. let the MIN be the minimum integer in T and MAX be the maximum, then the cost of set T if defined as (MAX-MIN) ^ 2. now given an integer set S, we want to find out M subsets S1, S2 ,..., SM of S, such that

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Sample Output

Case 1: 1Case 2: 18HintThe answer will fit into a 32-bit signed integer. 

Source2010 ACM-ICPC Multi-University Training Contest (5) -- Host by BJTU
A set containing N elements is divided into M subsets, so that the maximum and least squares of each subset are the same.

Idea: you can think of greedy sorting elements from small to large, and quickly get the minimum cost of j elements in the previous I subsets of dp [I] [j, dp equation dp [I] [j] = dp [I-1] [k] + (a [j]-a [k + 1]); but three loops are required, time complexity is unacceptable, so optimization is required. The slope optimization and quadrilateral inequality are acceptable. I am using the slope optimization just learned. Set k1 Code:

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             # Define maxn 205 # define MAXN 100005 # define mod 100000000 # define INF 0x3f3f3f # define pi acos (-1.0) # define eps 1e-6typedef long ll; using namespace std; int n, m, ans, cnt, tot, flag; int a [10005], dp [2] [10005]; int q [10005]; int sqr (int x) {return x * x;} int get (int I, int k) {return dp [I] [k] + a [k + 1] * a [k + 1];} void solve () {int I, j, t, dx1, dy1, dx2, dy2, k1, k2, turn = 0; int head, tail; sort (a + 1, a + n + 1); memset (dp, 0x3f, sizeof (dp); dp [0] [0] = 0; for (I = 1; I <= m; I ++) {head = 0; tail =-1; q [+ tail] = I-1; for (j = I; j <= n; j ++) {while (head