HDU

Problem Destribe

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 cost of each subset is minimal.

Input

The input contains multiple test cases.
In the first line of the input there’s an integer T which is the number of test cases. Then the description of T test cases will be given.
For any test case, the first line contains two integers N (≤ 10,000) and M (≤ 5,000). N is the number of elements in S (may be duplicated). M is the number of subsets that we want to get. In the next line, there will be N integers giving set S.

Output

For each test case, output one line containing exactly one integer, the minimal total cost. Take a look at the sample output for format.

Sample Input

2
3 2
1 2 4
4 2
4 7 10 1

Sample Output

Case 1: 1
Case 2: 18

Hint

The answer will fit into a 32-bit signed integer.

题意 : 把n个数字 分成m个集合。 每个集合的价值是 这个集合中 (max-min)^2。 输出最

少的价值

思路 : 我们可以使用 dp[i][j] 代表 以 j 结尾 第 j 个集合的最小值

那么可以容易的推出来 dp[i][j] = min { dp[k][j-1] + ( v[j] - v[k] ) * ( v[j] - v[k] ) } ;

但是这个算法的时间复杂度是 O( n^3 ) 很容易 TLE ，因此我们需要用到斜率进行优化 ，

其实这个题几乎是斜率优化的裸题了，没学过的可以学习一下

AC code :

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long ll;

const int maxn = 1e4+50;
const int maxm = 5e3+50;

ll dp[maxn][maxm] ,v[maxn] ;
int n ,m ;
int que[maxn] ,head ,tail ;

ll getdp(int i,int j,int k) {
	return dp[k][j-1] + (v[i] - v[k + 1]) * (v[i] - v[k + 1]);
}

ll y(int j,int k,int q) {
	return dp[k][j-1] + v[k + 1] * v[k + 1] - (dp[q][j-1] + v[q + 1] * v[q + 1]);
}

ll x(int k,int q) {
	return 2 * ( v[k + 1] - v[q + 1] );
}

int main() {
	int t ,ncase = 1 ; cin>>t;
	while(t--) {
		scanf("%d %d",&n ,&m );
		for (int i = 1;i<=n;i++) scanf("%lld",&v[i] );
		sort(v + 1 ,v + n + 1 );
		for (int i = 1;i<=n;i++) dp[i][1] = (v[i] - v[1]) * (v[i] - v[1]);
		for (int i = 2;i<=m;i++) {
			head = tail = 0; que[tail ++] = i - 1;
			for (int j = i;j<=n;j++) {
				while(head + 1 < tail && y(i ,que[head+1] ,que[head]) < x(que[head+1] ,que[head]) * v[j] ) head ++;
				dp[j][i] = getdp(j ,i ,que[head] );
				while(head + 1 < tail && y(i ,que[tail-1] ,que[tail-2]) * x(j ,que[tail-1]) >= y(i ,j ,que[tail-1]) * x(que[tail-1] ,que[tail-2]) ) tail --;
				que[tail ++] = j;
			}
		}
		printf("Case %d: ",ncase++);
		printf("%lld
",dp[n][m]);
	}
	return 0;
}