[HDU 1421] dormitory migration (innovative DP)

Question address: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1421

N items are given. You need to select K pairs of items to minimize the sum of the squares of each pair of items.

Thinking; Dynamic Planning and solving, first sort items by weight in ascending order. According to greedy ideas, each pair of items must be two adjacent items, so that the square of the weight difference can be minimized. Then, DP uses f [I] [J] to represent the first I item, and selects j pairs at the minimum cost. An important classification discussion is required here:

(1) If J * 2 = I, that is, all the first I items are used for pairing, it is obvious that J is for item No. I-1 and no. I, f [I] [J] = f [I-2] [J-1] + (W [I]-W [I-1]) ^ 2

(2) other cases. The first I items are not completely used for pairing. You can select the I item not to be paired. f [I] [J] = f [I-1] [J], or I items are paired, same as (1), I items are paired with I-1 items, f [I] [J] = f [I-2] [J-1] + (W [I]-W [I-1]) ^ 2

The final answer is f [N] [K].

# Include <iostream> # include <stdio. h> # include <stdlib. h> # include <string. h >#include <algorithm> # define maxn 2010 using namespace STD; int f [maxn] [maxn]; // F [I] [J] = previous I items, select the minimum cost of the J pair int W [maxn]; int min (int A, int B) {if (a <B) return a; return B;} int main () {int N, K; while (scanf ("% d", & N, & K )! = EOF) {memset (F, 0, sizeof (f); W [0] = 0; For (INT I = 1; I <= N; I ++) scanf ("% d", & W [I]); sort (W + 1, W + n + 1); For (INT I = 2; I <= N; I ++) // The first I items for (Int J = 1; j * 2 <= I; j ++) // select the J pair {if (I! = 2 * j) f [I] [J] = min (F [I-2] [J-1] + (W [I]-W [I-1]) * (W [I]-W [I-1]), f [I-1] [J]); // either the I-th item is paired with the I-1-th item, either keep the first item unpaired else f [I] [J] = f [I-2] [J-1] + (W [I]-W [I-1]) * (W [I]-W [I-1]);} printf ("% d \ n", F [N] [k]);} return 0 ;}

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[HDU 1421] dormitory migration (innovative DP)