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: Time Limit Exceeded)



import java.util.HashMap;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
public class Main {
    static final int MAX = Integer.MAX_VALUE;
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        HashMap<Integer, Integer>[] d = new HashMap[n+1];
        for (int i = 0; i <= n; i++) {
            d[i] = new HashMap<Integer, Integer>();
        }
        
        for (int i = 0; i < m; i++) {
            int xLi = sc.nextInt();
            int xRi = sc.nextInt();
            int xDi = sc.nextInt();
            d[xLi].put(xRi, xDi);
            d[xRi].put(xLi, -xDi);
        }

        
        Queue<Integer> q = new LinkedList<Integer>();
        long[] pos = new long[n+1];
        boolean[] visited = new boolean[n+1];
        
        for (int i = 1; i <= n; i++) {
            if (visited[i]) continue;
            
            q.add(i);
            pos[i] = 0;
            while (!q.isEmpty()) {
                int first = q.poll();
                Set<Integer> keys = d[first].keySet();
                for (int k : keys) {
                    Integer costFK = d[first].get(k);
//                    U.println(costFK);
                    if (costFK != null && costFK != MAX) {
                        d[first].put(k, MAX);
                        if (visited[k]) {
                            if (pos[k] != pos[first] + costFK) {
                                System.out.println("No");
                                return;
                            }
                        } else {
                            visited[k] = true;
                            pos[k] = pos[first] + costFK;
                        }
                        q.add(k);
                    }
                }
            }
        }
        System.out.println("Yes");
    }
}