# Moo Fest (POJ 3109)

A simple problem need using binary indexed tree. At first we sort the cows according to their volumes, then we iterate over all cows and calculate the volume produced by current cow and added them into a binary indexed tree according to their position. As the cows is sorted by volumes so for each cow i, previous cows added in the tree all have volumes less than or equal to its volume. So the volume of converse should between i with these cows is i's volume. We can use binary indexed tree to fast query the number of cows on the left or right of i and apply different formular to calculate the total distance according to their distances to the left most coordinate.

As the position space is large and we can not cost so much memory, we have to find all distinct x coordinate and compress them.

import java.io.*;
import java.util.*;

public class MooFest {
private static void add(long[] bit, int i, int x) {
while (i > 0 && i < bit.length) {
bit[i] += x;
i += i & -i;
}
}

private static long query(long[] bit, int i) {
long res = 0;
while (i > 0 && i < bit.length) {
res += bit[i];
i -= i & -i;
}
return res;
}

public static void main(String[] args) {
Scanner in = new Scanner(System.in);

while (in.hasNextInt()) {
int N = in.nextInt();
int[][] cows = new int[N];
for (int i = 0; i < N; i++) {
cows[i] = in.nextInt();
cows[i] = in.nextInt();
}

Arrays.sort(cows, new Comparator<int[]> () {
public int compare(int[] a, int[] b) {
return a - b;
}
});

long[] bitcount = new long;
long[] bitdistance = new long;

long res = 0;
for (int i = 0; i < N; i++) {
int vol = cows[i];
int x = cows[i];
long leftCount = query(bitcount, x);
long rightCount = i - leftCount;
long leftDistance = query(bitdistance, x);
long rightDistance = query(bitdistance, 20000) - leftDistance;
res += (leftCount * x - leftDistance + rightDistance - rightCount * x) * vol;