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.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int M = sc.nextInt();
		int[] L = new int[M];
		int[] R = new int[M];
		int[] D = new int[M];
		for(int i = 0; i < M; i++) {
			L[i] = sc.nextInt();
			R[i] = sc.nextInt();
			D[i] = sc.nextInt();
		}
		int[] x = new int[N + 1];
		for(int i = 1; i < N + 1; i++) {
			x[i] = -1;
		}
		boolean ans = true;
		for(int i = 0; i < M; i++) {
			if((x[L[i]] == -1) && (x[R[i]] == -1)) {
				x[L[i]] = 0;
				x[R[i]] = D[i];
			} else if((x[L[i]] == -1) && !(x[R[i]] == -1)) {
				if(x[R[i]] > D[i]) {
					x[L[i]] = x[R[i]] - D[i];
				} else {
					for(int j = 1; j < N + 1; j++) {
						if(!(x[j] == -1)) {
							x[j] += D[i] - x[R[i]];
						}
					}
					x[L[i]] = 0;
				}
			} else if(!(x[L[i]] == -1) && (x[R[i]] == -1)) {
				x[R[i]] = x[L[i]] + D[i];
			} else {
				if(x[R[i]] - x[L[i]] != D[i]) {
					ans = false;
					break;
				}
			}
		}
		if(ans) {
			System.out.println("Yes");
		} else {
			System.out.println("No");
		}
	}

}