解题思路
第一步显然是将原数组排序嘛……然后分成一些不相交的子集，这样显然最小。重点是怎么分。
首先，我们写出一个最暴力的\(DP\)：
我们令$F[ i ][ j ] $ 为到第\(i\)位，分成\(j\)组的代价，我们可以写出如下 $ DP$
for( LL i = 1; i <= N; ++i ) F[ i ][ 1 ] = sqr( A[ i ] - A[ 1 ] );
for( LL j = 2; j <= M; ++j ) for( LL i = j; i <= N; ++i ) {F[ i ][ j ] = INF;for( LL k = j - 1; k <= i - 1; ++k ) F[ i ][ j ] = min( F[ i ][ j ], F[ k ][ j - 1 ] + sqr( A[ j ] - A[ k + 1 ] ) );}滚动掉一维节省空间：
for( LL i = 1; i <= N; ++i ) F1[ i ] = sqr( A[ i ] - A[ 1 ] );
for( LL j = 2; j <= M; ++j ) {for( LL i = j; i <= N; ++i ) {F2[ i ] = INF;for( LL k = j - 1; k <= i - 1; ++k )F2[ i ] = min( F2[ i ], F1[ K ] + sqr( A[ j ] - A[ k + 1 ] ) );}memcpy( F1, F2, sizeof( F2 ) );
}但是时间上依然难以接受。我们考虑优化最内层的转移复杂度。
不妨设转移的时候\(l>k\)且从\(l\)转移优于从\(k\)转移，那么我们就能得到如下不等式：
\[ F_1[l]+(A[j]-A[l+1])^2<F_1[k]+(A[j]-A[k+1])^2 \]
化简，得
\[ \frac{F_1[l]-F_1[k]+A[l+1]^2-A[k+1]^2}{2A[l+1]-2A[k+1]}<A[j] \]
所以我们可以进行斜率优化。具体优化讲解可以看这里。过程十分相似。
参考程序
#include <bits/stdc++.h>
#define LL long long
using namespace std;LL N, M, A[ 10010 ], F1[ 10010 ], F2[ 10010 ];
LL L, R, Queue[ 10010 ];inline LL sqr( LL x ) { return x * x; }inline void Clear() {memset( A, 0, sizeof( A ) );memset( F1, 0, sizeof( F1 ) );return;
}inline void Clear2() {memset( Queue, 0, sizeof( Queue ) );memset( F2, 0, sizeof( F2 ) );L = R = 0;return;
}inline bool Less( LL i, LL j, LL T ) {LL DeltaX = 2 * ( A[ j + 1 ] - A[ i + 1 ] );LL DeltaY = F1[ j ] - F1[ i ] + sqr( A[ j + 1 ] ) - sqr( A[ i + 1 ] );return DeltaY < T * DeltaX;
}inline bool Greater( LL i, LL j, LL k ) {LL DeltaX1 = 2 * ( A[ j + 1 ] - A[ i + 1 ] );LL DeltaX2 = 2 * ( A[ k + 1 ] - A[ j + 1 ] );LL DeltaY1 = F1[ j ] - F1[ i ] + sqr( A[ j + 1 ] ) - sqr( A[ i + 1 ] );LL DeltaY2 = F1[ k ] - F1[ j ] + sqr( A[ k + 1 ] ) - sqr( A[ j + 1 ] );return DeltaX2 * DeltaY1 >= DeltaX1 * DeltaY2;
}void Work() {Clear();scanf( "%lld%lld", &N, &M );for( LL i = 1; i <= N; ++i ) scanf( "%lld", &A[ i ] );sort( A + 1, A + N + 1, less< LL >() );for( LL i = 1; i <= N; ++i ) F1[ i ] = sqr( A[ i ] - A[ 1 ] );for( LL j = 2; j <= M; ++j ) {Clear2();Queue[ R++ ] = j - 1;for( LL i = j; i <= N; ++i ) {while( L + 1 < R && Less( Queue[ L ], Queue[ L + 1 ], A[ i ] ) )++L;F2[ i ] = F1[ Queue[ L ] ] + sqr( A[ i ] - A[ Queue[ L ] + 1 ] );while( L + 1 < R && Greater( Queue[ R - 2 ], Queue[ R - 1 ], i ) )--R;Queue[ R++ ] = i;}memcpy( F1, F2, sizeof( F2 ) );}printf( "%lld\n", F1[ N ] );return;
}int main() {LL t; scanf( "%lld", &t );for( LL i = 1; i <= t; ++i ) {printf( "Case %lld: ", i );Work();}return 0;
}
