Fair Cut
Problem Statement :
Li and Lu have n integers, a1,a2, ...,an, that they want to divide fairly between the two of them. They decide that if Li gets integers with indices I = {i1, i2, ..., ik} (which implies that Lu gets integers with indices J = {1, 2, ...,n}\I), then the measure of unfairness of this division is:
f(I) = Σ(i∈I) Σ(j∈J)|ai-aj|
Find the minimum measure of unfairness that can be obtained with some division of the set of integers where Li gets exactly k integers.
Note A\B means Set complement
Input Format
The first line contains two space-separated integers denoting the respective values of n(the number of integers Li and Lu have) and k (the number of integers Li wants).
The second line contains n space-separated integers describing the respective values of a1, a2, ..., an.
Constraints
1 <= k < n < =3000
1 <= ai <= 10^9
For 15% of the test cases, n <=20.
For 45% of the test cases, n <=40.
Output Format
Print a single integer denoting the minimum measure of unfairness of some division where Li gets k integers.
Solution :
Solution in C :
In C++ :
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
const ll Inf = 4000000000000000000;
const int Maxn = 3005;
int n, k;
int a[Maxn];
ll dp[Maxn][Maxn];
int main() {
scanf("%d %d", &n, &k);
for (int i = 0; i < n; i++)
scanf("%d", &a[i]);
sort(a, a + n);
fill((ll*)dp, (ll*)dp + Maxn * Maxn, Inf);
dp[0][0] = 0;
for (int i = 0; i < n; i++)
for (int j = 0; j <= i && j <= k; j++) if (dp[i][j] < Inf) {
int my = j, his = i - j;
if (my < k) {
ll add = ll(a[i]) * (his - (n - k - his));
dp[i + 1][j + 1] = min(dp[i + 1][j + 1], dp[i][j] + add);
}
if (his < n - k) {
ll add = ll(a[i]) * (my - (k - my));
dp[i + 1][j] = min(dp[i + 1][j], dp[i][j] + add);
}
}
printf("%lld\n", dp[n][k]);
return 0;
}
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
private static long[] DIFF;
private static long[] DATA;
private static int COUNT;
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
COUNT = sc.nextInt();
DATA = new long[n];
for(int i = 0 ; i < n ; i++) {
DATA[i] = sc.nextLong();
}
COUNT = Math.min(DATA.length - COUNT, COUNT);
DIFF = new long[DATA.length];
System.out.println(calc2());
}
private static long calc2() {
Arrays.sort(DATA);
long[] sum = new long[DATA.length + 1];
for (int i = 1; i <= DATA.length; i++) {
sum[i] = sum[i - 1] + DATA[i - 1];
}
for (int i = 0; i < DATA.length; i++) {
for (long aT : DATA) {
DIFF[i] += Math.abs(DATA[i] - aT);
}
}
return Math.min(clc(0), clc(1));
}
private static long clc(int i) {
long chSum = 0;
long summ0 = 0;
for (int j = 0; j < DATA.length; j++) {
if (!valid(j, i)) {
continue;
}
chSum += DATA[j];
for (int m = 0; m < DATA.length; m++) {
if (!valid(m, i)) {
summ0 += Math.abs(DATA[j] - DATA[m]);
}
}
}
long min = summ0;
int first = i;
int next = 2 * COUNT + i;
while (next < DATA.length) {
chSum -= DATA[first];
summ0 -= DIFF[first];
summ0 += 2 * (chSum - DATA[first] * (COUNT - 1));
summ0 += DIFF[next];
summ0 -= 2 * (DATA[next] * (COUNT - 1) - chSum);
chSum += DATA[next];
min = Math.min(summ0, min);
next += 2;
first += 2;
}
return min;
}
private static boolean valid(int j, int i) {
return j % 2 == i && j < COUNT * 2;
}
}
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int n;
int k;
int a[3001];
long long listcost[3001];
long long allcost[3001];
int main()
{
int i;
int j;
int besti;
long long bestscore;
long long allscore;
long long tmp;
scanf("%d %d", &n, &k);
if (k > n - k) {
k = n - k;
}
for (i = 0; i < n; ++i) {
scanf("%d", &a[i]);
}
for (i = 1; i < n; ++i) {
tmp = a[i];
j = i;
while (j) {
if (tmp > a[j-1]) {
break;
}
a[j] = a[j-1];
j -= 1;
}
a[j] = tmp;
}
for (i = 0; i < n; ++i) {
listcost[i] = 0;
allcost[i] = 0;
}
for (i = 0; i < n; ++i) {
for (j = 0; j < n; ++j) {
if (a[i] > a[j]) {
allcost[i] += a[i] - a[j];
} else {
allcost[i] += a[j] - a[i];
}
}
}
allscore = 0;
bestscore = 0;
for (i = 0; i < k; ++i) {
allscore = bestscore;
besti = 0;
bestscore = 1000LL*1000LL*1000LL*1000LL*1000LL;
for (j = 0; j < n; ++j) {
tmp = allcost[j];
tmp -= listcost[j];
tmp -= listcost[j];
tmp += allscore;
if (tmp < bestscore) {
bestscore = tmp;
besti = j;
}
if (tmp == bestscore) {
if (listcost[j] > listcost[besti]) {
besti = j;
}
}
}
tmp = allcost[besti];
allcost[besti] = allcost[n - 1];
allcost[n - 1] = tmp;
tmp = listcost[besti];
listcost[besti] = listcost[n - 1];
listcost[n - 1] = tmp;
tmp = a[besti];
a[besti] = a[n - 1];
a[n - 1] = (int) tmp;
n -= 1;
j = besti;
while (j < n - 1) {
if (a[j] < a[j+1]) {
break;
}
tmp = allcost[j];
allcost[j] = allcost[j+1];
allcost[j+1] = tmp;
tmp = listcost[j];
listcost[j] = listcost[j+1];
listcost[j+1] = tmp;
tmp = a[j];
a[j] = a[j+1];
a[j+1] = (int) tmp;
j += 1;
}
// update listcost
for (j = 0; j < n; ++j) {
if (a[j] > a[n]) {
listcost[j] += a[j] - a[n];
} else {
listcost[j] += a[n] - a[j];
}
}
}
printf("%lld\n", bestscore);
return 0;
}
In python3 :
import sys
def solve(a, k):
def sumup(I, J):
sm = 0
for u in I:
for v in J:
sm += abs(u - v)
return sm
k = min(k, len(a) - k)
a.sort()
print('k=%d a=%s' % (k, a), file = sys.stderr)
index = (len(a) - 2 * k) // 2
I = a[index:index + 2 * k:2]
J = a[:index] + a[index + 1:index + 2 * k + 1:2] + a[index + 2 * k:]
print('I=%s J=%s' % (I, J), file = sys.stderr)
return sumup(I, J)
_, k = map(int, input().split())
a = list(map(int, input().split()))
print(solve(a, k))
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