Problem:
Score : 900 points

Problem Statement

In the city of Nevermore, there are 10^8 streets and 10^8 avenues, both numbered from 0 to 10^8-1.
All streets run straight from west to east, and all avenues run straight from south to north.
The distance between neighboring streets and between neighboring avenues is exactly 100 meters.

Every street intersects every avenue. Every intersection can be described by pair (x, y), where x is avenue ID and y is street ID.

There are N fountains in the city, situated at intersections (X_i, Y_i).
Unlike normal intersections, there's a circle with radius 10 meters centered at the intersection, and there are no road parts inside this circle.

The picture below shows an example of how a part of the city with roads and fountains may look like.

City governors don't like encountering more than one fountain while moving along the same road.
Therefore, every street contains at most one fountain on it, as well as every avenue.

Citizens can move along streets, avenues and fountain perimeters.
What is the shortest distance one needs to cover in order to get from intersection (x_1, y_1) to intersection (x_2, y_2)?

Constraints

0 \leq x_1, y_1, x_2, y_2 < 10^8

1 \leq N \leq 200,000

0 \leq X_i, Y_i < 10^8

X_i \neq X_j for i \neq j

Y_i \neq Y_j for i \neq j

Intersections (x_1, y_1) and (x_2, y_2) are different and don't contain fountains.

All input values are integers.

Input

Input is given from Standard Input in the following format:

x_1 y_1 x_2 y_2
N
X_1 Y_1
X_2 Y_2
:
X_N Y_N

Output

Print the shortest possible distance one needs to cover in order to get from intersection (x_1, y_1) to intersection (x_2, y_2), in meters.
Your answer will be considered correct if its absolute or relative error doesn't exceed 10^{-11}.

Sample Input 1

1 1 6 5
3
3 2
5 3
2 4

Sample Output 1

891.415926535897938

One possible shortest path is shown on the picture below. The path starts at the blue point, finishes at the purple point and follows along the red line.

Sample Input 2

3 5 6 4
3
3 2
5 3
2 4

Sample Output 2

400.000000000000000

Sample Input 3

4 2 2 2
3
3 2
5 3
2 4

Sample Output 3

211.415926535897938

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
import java.util.stream.Collectors;
import java.io.*;

public class Main {

	int MAX = 100_000_000;

	class Point {
		int x, y;

		public Point(int x, int y) {
			this.x = x;
			this.y = y;
		}

		@Override
		public String toString() {
			return "Point [x=" + x + ", y=" + y + "]";
		}

	}

	void solve() {
		List<Point> pts = new ArrayList<>();
		pts.add(new Point(in.nextInt(), in.nextInt()));
		pts.add(new Point(in.nextInt(), in.nextInt()));
		int n = in.nextInt();
		for (int i = 0; i < n; i++) {
			pts.add(new Point(in.nextInt(), in.nextInt()));
		}

		Point start = pts.get(0);
		Point finish = pts.get(1);
		if (start.x > finish.x) {
			for (Point p : pts) {
				p.x = MAX - 1 - p.x;
			}
		}
		if (start.y > finish.y) {
			for (Point p : pts) {
				p.y = MAX - 1 - p.y;
			}
		}

		double result = 100.0 * (0.0 + finish.x + finish.y - start.x - start.y);
		pts = pts.stream().filter(p -> start.x <= p.x && p.x <= finish.x && start.y <= p.y && p.y <= finish.y)
				.collect(Collectors.toList());
		n = pts.size();
		int[] x = new int[n], y = new int[n];
		for (int i = 0; i < n; i++) {
			x[i] = pts.get(i).x;
			y[i] = pts.get(i).y;
		}
		x = removeDupes(x);
		y = removeDupes(y);
		for (int i = 0; i < n; i++) {
			pts.get(i).x = Arrays.binarySearch(x, pts.get(i).x);
			pts.get(i).y = Arrays.binarySearch(y, pts.get(i).y);
		}

		Collections.sort(pts, (o1, o2) -> {
			if (o1.x != o2.x) {
				return o1.x - o2.x;
			}
			return o1.y - o2.y;
		});

		SegmentTree tree = new SegmentTree(y.length + 1);
		if (pts.get(1).x == start.x) {
			tree.set(0, 0);
		} else {
			tree.updateMax(0, y.length, 0);
		}
		
		for (int i = 1; i < pts.size() - 1; i++) {
			double down = tree.get(0, pts.get(i).y - 1);
			double same = tree.get(pts.get(i).y, pts.get(i).y);
			{ // update same
				double cur1 = down + (20 - 5 * Math.PI);
				double cur2 = same + (20 - 10 * Math.PI);
				tree.set(pts.get(i).y, Math.max(cur1, cur2));
			}
			{ // update up
				double cur1 = same + 20 - 5 * Math.PI;
				double cur2 = down + 20 - 10 * Math.PI;
				tree.updateMax(pts.get(i).y + 1, y.length, Math.max(cur1, cur2));
			}
		}

		result -= tree.get(finish.y, finish.y);
		out.println(result);
	}

	int[] removeDupes(int[] x) {
		Arrays.sort(x);
		int cnt = 0;
		for (int i = 0; i < x.length; i++) {
			if (i == 0 || x[i] > x[i - 1]) {
				cnt++;
			}
		}
		int[] result = new int[cnt];
		for (int i = 0, j = 0; i < x.length; i++) {
			if (i == 0 || x[i] != x[i - 1]) {
				result[j++] = x[i];
			}
		}
		return result;
	}

	double INF = Integer.MIN_VALUE;

	class SegmentTree {
		double[] max;
		double[] push;
		int size;

		public SegmentTree(int n) {
			max = new double[4 * n];
			push = new double[4 * n];
			size = n;
			Arrays.fill(max, INF);
			Arrays.fill(push, INF);
		}

		void set(int pos, double val) {
			set(0, size, pos, val, 0);
		}

		void set(int l, int r, int pos, double val, int i) {
			if (l + 1 == r) {
				max[i] = val;
				return;
			}
			push(i);
			int m = (l + r) >> 1;
			if (pos < m) {
				set(l, m, pos, val, 2 * i + 1);
			} else {
				set(m, r, pos, val, 2 * i + 2);
			}
			refresh(i);
		}

		double get(int l, int r) {
			return get(0, size, l, r + 1, 0);
		}

		double get(int l, int r, int left, int right, int i) {
			if (left <= l && r <= right) {
				return max[i];
			} else if (right <= l || r <= left) {
				return INF;
			} else {
				push(i);
				int m = (l + r) >> 1;
				double ret = Math.max(get(l, m, left, right, 2 * i + 1), get(m, r, left, right, 2 * i + 2));
				refresh(i);
				return ret;
			}
		}

		void updateMax(int from, int to, double val) {
			updateMax(0, size, from, to + 1, val, 0);
		}

		void push(int i) {
			if (push[i] != INF) {
				updateOne(2 * i + 1, push[i]);
				updateOne(2 * i + 2, push[i]);
				push[i] = INF;
			}
		}

		void updateOne(int i, double value) {
			max[i] = Math.max(max[i], value);
			push[i] = Math.max(push[i], value);
		}

		void refresh(int i) {
			max[i] = Math.max(max[2 * i + 1], max[2 * i + 2]);
		}

		void updateMax(int l, int r, int left, int right, double value, int i) {
			if (left <= l && r <= right) {
				updateOne(i, value);
			} else if (right <= l || r <= left) {
				return;
			} else {
				push(i);
				int m = (l + r) >> 1;
				updateMax(l, m, left, right, value, 2 * i + 1);
				updateMax(m, r, left, right, value, 2 * i + 2);
				refresh(i);
			}
		}
	}

	FastScanner in;
	PrintWriter out;

	void run() {
		in = new FastScanner();
		out = new PrintWriter(System.out);
		solve();
		out.close();
	}

	class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		public FastScanner(String s) {
			try {
				br = new BufferedReader(new FileReader(s));
			} catch (FileNotFoundException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
		}

		public String nextToken() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
				}
			}
			return st.nextToken();
		}

		public int nextInt() {
			return Integer.parseInt(nextToken());
		}

		public long nextLong() {
			return Long.parseLong(nextToken());
		}

		public double nextDouble() {
			return Double.parseDouble(nextToken());
		}
	}

	public static void main(String[] args) {
		new Main().run();
	}
}