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 {

    int N;
    int M;
    ArrayList<Integer>[] g;
    boolean[] used;
    boolean[] qued;
    int[] dis;
    int INF = Integer.MAX_VALUE;

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

    void solve() {
        Scanner sc = new Scanner(System.in);
        N = sc.nextInt();
        M = sc.nextInt();
        g = new ArrayList[N];
        for (int i = 0; i < N; i++) {
            g[i] = new ArrayList<Integer>();
        }
        used = new boolean[N];
        qued = new boolean[N];
        Arrays.fill(used, false);
        Arrays.fill(qued, false);
        for (int i = 0; i < M; i++) {
            int L = sc.nextInt();
            int R = sc.nextInt();
            int D = sc.nextInt();
            g[L - 1].add(R - 1);
            g[L - 1].add(D);
            g[R - 1].add(L - 1);
            g[R - 1].add(D * -1);
        }
        dis = new int[N];
        for (int i = 0; i < N; i++) {
            dis[i] = INF;
        }
        boolean ans = true;
        for (int i = 0; i < N; i++) {
            if (!used[i]) {
                boolean res = dfs(i);
                if (!res) {
                    ans = false;
                }
            }
        }
        System.out.println(ans ? "Yes" : "No");
    }

    boolean dfs(int v) {
        boolean res = true;
        Deque<Integer> q = new ArrayDeque<Integer>();
        q.push(v);
        dis[v] = 0;
        while (q.size() != 0) {
            int now = q.pop();
            used[now] = true;
            for (int i = 0; i < g[now].size(); i += 2) {
                int next = g[now].get(i);
                if (!used[next]) {
                    dis[next] = dis[now] + g[now].get(i + 1);
                } else if (dis[next] != dis[now] + g[now].get(i + 1)) {
                    res = false;
                }
                if (!qued[next]) {
                    q.push(next);
                    qued[next] = true;
                }
            }
        }
        return res;
    }
}