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

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.

題意 : 把n個數字 分成m個集合。 每個集合的價值是 這個集合中 (max-min)^2。 輸出最

少的價值

思路 : 我們可以使用 dp[i][j] 代表 以 j 結尾 第 j 個集合的最小值

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

但是這個演算法的時間複雜度是 O( n^3 ) 很容易 TLE ，因此我們需要用到斜率進行優化 ，

其實這個題幾乎是斜率優化的裸題了，沒學過的可以學習一下

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\n",dp[n][m]);
	}
	return 0;
}