Division Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 999999/400000 K (Java/Others)
Total Submission(s): 2676 Accepted Submission(s): 1056
Problem Description Little 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
and the total cZ??http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vc3Qgb2YgZWFjaCBzdWJzZXQgaXMgbWluaW1hbC4gCiAKPGJyPgpJbnB1dApUaGUgaW5wdXQgY29udGFpbnMgbXVsdGlwbGUgdGVzdCBjYXNlcy48YnI+CkluIHRoZSBmaXJzdCBsaW5lIG9mIHRoZSBpbnB1dCB0aGVyZaGvcyBhbiBpbnRlZ2VyIFQgd2hpY2ggaXMgdGhlIG51bWJlciBvZiB0ZXN0IGNhc2VzLiBUaGVuIHRoZSBkZXNjcmlwdGlvbiBvZiBUIHRlc3QgY2FzZXMgd2lsbCBiZSBnaXZlbi4KPGJyPgpGb3IgYW55IHRlc3QgY2FzZSwgdGhlIGZpcnN0IGxpbmUgY29udGFpbnMgdHdvIGludGVnZXJzIE4gKKHcIDEwLDAwMCkgYW5kIE0gKKHcIDUsMDAwKS4gTiBpcyB0aGUgbnVtYmVyIG9mIGVsZW1lbnRzIGluIFMgKG1heSBiZSBkdXBsaWNhdGVkKS4gTSBpcyB0aGUgbnVtYmVyIG9mIHN1YnNldHMgdGhhdCB3ZSB3YW50IHRvIGdldC4gSW4gdGhlIG5leHQgbGluZSwgdGhlcmUgd2lsbCBiZSBOIGludGVnZXJzIGdpdmluZyBzZXQgUy48YnI+Cjxicj4KCiAKPGJyPgpPdXRwdXQKRm9yIGVhY2ggdGVzdCBjYXNlLCBvdXRwdXQgb25lIGxpbmUgY29udGFpbmluZyBleGFjdGx5IG9uZSBpbnRlZ2VyLCB0aGUgbWluaW1hbCB0b3RhbCBjb3N0LiBUYWtlIGEgbG9vayBhdCB0aGUgc2FtcGxlIG91dHB1dCBmb3IgZm9ybWF0Ljxicj4KPGJyPgoKIAo8YnI+ClNhbXBsZSBJbnB1dAoKPHByZSBjbGFzcz0="brush:java;">2 3 2 1 2 4 4 2 4 7 10 1
Sample Output
Case 1: 1
Case 2: 18
HintThe answer will fit into a 32-bit signed integer.
Source 2010 ACM-ICPC Multi-University Training Contest（5）――Host by BJTU
题意： 将含有N个元素的一个集合分成M个子集，使得每个子集的最大值与最小值平方差的和最小。
思路： 可以想到贪心将元素从小到大排序，很快可以得出dp[i][j]-前i个子集分j个元素的最小花费， dp方程 dp[i][j]=dp[i-1][k]+(a[j]-a[k+1])；但是需要三重循环，时间复杂度接受不了，所以需要优化，这题斜率优化和四边形不等式都可以，我采用的是刚学习的斜率优化。 设k1
代码：
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