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.*;
import static java.lang.System.*;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		D2(sc);
	}
	

	public static void D2(Scanner sc) {
		int N = sc.nextInt();
		int M = sc.nextInt();
		HashMap<Integer,Integer>[] edge = new HashMap[N]; // edge[a].get(b) はノードaからノードbへの距離   
		boolean[] visit = new boolean[N];
		long[] value = new long[N];
		for(int i=0;i<N;i++) edge[i] = new HashMap<Integer,Integer>();
		for(int i=0;i<M;i++) {
			int l = sc.nextInt()-1;
			int r = sc.nextInt()-1;
			int d = sc.nextInt();
			if(!putedge(edge,l,r,d)) {
				out.println("No");
				return;
			}
		}
		for(int i=0;i<N;i++) {
			if(!bfs(edge,i,value,visit)) {
				out.println("No");
				return;
			};
		}
		out.println("Yes");
	}
	
	public static boolean putedge(HashMap<Integer,Integer>[] edge, int l, int r, int d) {
		if(edge[l].containsKey(r) && edge[l].get(r)!=d) return false;
		edge[l].put(r,d);
		edge[r].put(l,-d);
		return true;
	}
	// BFSでvalueを計算して行く。　valueが矛盾すればfalse
	public static boolean bfs(HashMap<Integer,Integer>[] edge, int n, long[] value, boolean[] visit) {
		if(visit[n]) return true;
		LinkedList<Integer> que = new LinkedList<Integer>();
		visit[n] = true;
		value[n] = 0;
		que.add(n);
		while(!que.isEmpty()) {
			int node = que.removeFirst();
			for(int neigh : edge[node].keySet()) {
				if(visit[neigh] && value[neigh]!=(value[node]+edge[node].get(neigh))) {
					return false;
				}
				else if(!visit[neigh]){
					visit[neigh] = true;
					value[neigh] = value[node]+edge[node].get(neigh);
					que.add(neigh);
 				}
			}
		}
		return true;
	}
	
}