# USACO 2014 December Contest, Bronze Problem 1. Marathon

USACO2014-DEC-B1

(Analysis by Nick Wu)

Our first instinct when trying to figure out which point to skip is to try all of them. If we choose to skip each point and compute the new distance directly, then it takes about NN operations to compute the distance and there are about NN points to check, giving us an algorithm which runs in about NNoperations. This will be too slow when NN gets to be 100,000.

Let's take a closer look at what happens when you skip a specific point. If we number the points from 1 to NN, and skip point KK, then the path we take goes from point 1 to point K1K−1, then from point K1K−1 to point K+1K+1, and then from point K+1K+1 to point NN. The distance of this path is exactly equal to the following:

(total distance without skipping any points) - (distance between points K1K−1 and KK) - (distance between points KK and K+1K+1) + (distance between points K1K−1 and K+1K+1).

If we compute the total distance without skipping any points beforehand, then figuring out how long the path is when we want to skip a specific point no longer requires NN operations! It only requires a constant number of operations, and that will be fast enough.

Here is my Java code:

import java.io.*;
import java.util.*;
public class marathon {
public static void main(String[] args) throws IOException {
PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("marathon.out")));
int[] x = new int[n];
int[] y = new int[n];
for(int i = 0; i < n; i++) {
x[i] = Integer.parseInt(st.nextToken());
y[i] = Integer.parseInt(st.nextToken());
}
int totalDistance = 0;
for(int i = 1; i < n; i++) {
totalDistance += Math.abs(x[i] - x[i-1]) + Math.abs(y[i] - y[i-1]);
}
int largestSkip = 0;
for(int i = 1; i < n-1; i++) {
int noSkipDistance = Math.abs(x[i+1] - x[i]) + Math.abs(x[i] - x[i-1]) + Math.abs(y[i+1] - y[i]) + Math.abs(y[i] - y[i-1]);
int skipDistance = Math.abs(x[i+1] - x[i-1]) + Math.abs(y[i+1] - y[i-1]);
largestSkip = Math.max(largestSkip, noSkipDistance - skipDistance);
}
pw.println(totalDistance - largestSkip);
pw.close();
}
}