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: Accepted)

import java.util.*;
public class Main {
    static int[] v;
    static int[] dist;
    static int INF = 1000000000;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        // List<Edge>[] g = new List[n + 1];
        List<List<Edge>> g = new ArrayList<>();
        for (int i = 0; i <= n; i++) {
            g.add(new ArrayList<>());
        }
        dist = new int[n + 1];
        for (int i = 0; i <= n; i++) {
            dist[i] = INF;
        }
        for (int i = 0; i < m; i++) {
            int l = sc.nextInt();
            int r = sc.nextInt();
            int d = sc.nextInt();
            // g[l].add(new Edge(r, d));
            // g[r].add(new Edge(l, -d));
            g.get(l).add(new Edge(r, d));
            g.get(r).add(new Edge(l, -d));
        }
        if (bfs(g, n)) {
            System.out.println("Yes");
            return;
        }
        System.out.println("No");
    }

    public static boolean bfs(List<List<Edge>> g, int n) {
        for (int i = 1; i <= n; i++) {
            if (dist[i] != INF) {
                continue;
            }
            Queue<Integer> que = new LinkedList<>();
            que.offer(i);
            dist[i] = 0;
            while (!que.isEmpty()) {
                int v = que.poll();
                for (Edge e : g.get(v)) {
                    if (dist[e.to] == INF) {
                        dist[e.to] = dist[v] + e.cost;
                        que.offer(e.to);
                    } else if (dist[e.to] != dist[v] + e.cost) {
                        return false;
                    }
                }
            }
        }
        return true;
    }
}
class Edge {
    int to;
    int cost;
    public Edge(int to, int cost) {
        this.to = to;
        this.cost = cost;
    }
}