Problem:
Score : 400 points

Problem Statement

There are N people standing on the x-axis.
Let the coordinate of Person i be x_i.
For every i, x_i is an integer between 0 and 10^9 (inclusive).
It is possible that more than one person is standing at the same coordinate.

You will given M pieces of information regarding the positions of these people.
The i-th piece of information has the form (L_i, R_i, D_i).
This means that Person R_i is to the right of Person L_i by D_i units of distance, that is, x_{R_i} - x_{L_i} = D_i holds.

It turns out that some of these M pieces of information may be incorrect.
Determine if there exists a set of values (x_1, x_2, ..., x_N) that is consistent with the given pieces of information.

Constraints

1 \leq N \leq 100 000

0 \leq M \leq 200 000

1 \leq L_i, R_i \leq N (1 \leq i \leq M)

0 \leq D_i \leq 10 000 (1 \leq i \leq M)

L_i \neq R_i (1 \leq i \leq M)

If i \neq j, then (L_i, R_i) \neq (L_j, R_j) and (L_i, R_i) \neq (R_j, L_j).

D_i are integers.

Input

Input is given from Standard Input in the following format:

N M
L_1 R_1 D_1
L_2 R_2 D_2
:
L_M R_M D_M

Output

If there exists a set of values (x_1, x_2, ..., x_N) that is consistent with all given pieces of information, print Yes; if it does not exist, print No.

Sample Input 1

3 3
1 2 1
2 3 1
1 3 2

Sample Output 1

Yes

Some possible sets of values (x_1, x_2, x_3) are (0, 1, 2) and (101, 102, 103).

Sample Input 2

3 3
1 2 1
2 3 1
1 3 5

Sample Output 2

No

If the first two pieces of information are correct, x_3 - x_1 = 2 holds, which is contradictory to the last piece of information.

Sample Input 3

4 3
2 1 1
2 3 5
3 4 2

Sample Output 3

Yes

Sample Input 4

10 3
8 7 100
7 9 100
9 8 100

Sample Output 4

No

Sample Input 5

100 0

Sample Output 5

Yes

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


import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {

	public static long[] list;
	public static List<Pair> info = new ArrayList<>();
	public static boolean ans = true;

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		list = new long[n];
		for(int i = 0; i < n; i++)
			list[i] = Integer.MAX_VALUE;
		for(int i = 0; i < m; i++) {
			int l = sc.nextInt() - 1;
			int r = sc.nextInt() - 1;
			long d = sc.nextLong();
			info.add(new Pair(l, r, d));
			info.add(new Pair(r, l, -d));
		}
		for(int i = 0; i < n; i++) {
			if(list[i] == Integer.MAX_VALUE) {
				list[i] = 0;
				Calc(i);
			}
		}
		if(ans) {
			System.out.println("Yes");
		} else {
			System.out.println("No");
		}
	}

	public static void Calc(int n) {
		for(Pair e : info) {
			if(e.getL() == n) {
				if(list[e.getR()] == Integer.MAX_VALUE) {
					list[e.getR()] = list[e.getL()] + e.getD();
					Calc(e.getR());
				} else {
					if(list[e.getR()] == list[e.getL()] + e.getD()) {
						ans = false;
						break;
					}
				}
			}
		}
	}

}

class Pair {
	int l;
	int r;
	long d;

	protected Pair(int l, int r, long d) {
		super();
		this.l = l;
		this.r = r;
		this.d = d;
	}

	public int getL() {
		return l;
	}

	public void setL(int l) {
		this.l = l;
	}

	public int getR() {
		return r;
	}

	public void setR(int r) {
		this.r = r;
	}

	public long getD() {
		return d;
	}

	public void setD(long d) {
		this.d = d;
	}

}