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.*;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        DJSet dj = new DJSet(n);
        int[][] e = new int[m][3];
        for (int i = 0; i < m; i++) {
            int l = sc.nextInt()-1;
            int r = sc.nextInt()-1;
            int d = sc.nextInt();
            e[i][0] = l; e[i][1] = r; e[i][2] = d;
            dj.union(l, r);
        }
        int[][][] g = buildWeightedGraph(n, e);
        Set<Integer> set = new HashSet<>();
        for (int i = 0; i < n; i++) {
            int r = dj.root(i);
            if(r < 0) r = i;
            set.add(r);
        }
        int[] p = new int[n];
        boolean[] used = new boolean[n];
        boolean ans = true;
        for(int r : set){
            ArrayDeque<Integer> q = new ArrayDeque<>();
            q.add(r);
            used[r] = true;
            while(!q.isEmpty()){
                int v = q.poll();
                used[v] = true;
                for(int[] na : g[v]){
                    int nv = na[0], nd = na[1];
                    if(used[nv]){
                        if(p[nv] != p[v] + nd){
                            ans = false;
                            break;
                        }
                    }else{
                        p[nv] = p[v] + nd;
                        q.add(nv);
                    }
                }
                if(!ans)break;
            }
            if(!ans)break;
        }
        System.out.println(ans ? "Yes" : "No");
        sc.close();

    }

    public static int[][][] buildWeightedGraph(int n, int[][] edges){
        int m = edges.length;
        int[][][] graph = new int[n][][];
        int[] deg = new int[n];
        for (int i = 0; i < m; i++) {
            deg[ edges[i][0] ]++;
            deg[ edges[i][1] ]++;
        }
        for (int i = 0; i < n; i++) {
            graph[i] = new int[deg[i]][2];
        }
        for (int i = 0; i < m; i++) {
            int a = edges[i][0];
            int b = edges[i][1];
            int c = edges[i][2];
            graph[a][--deg[a]][0] = b;
            graph[a][deg[a]][1] = c;
            graph[b][--deg[b]][0] = a;
            graph[b][deg[b]][1] -= c;
        }

        return graph;
    }

}

class DJSet {
    /**
     * 各要素の親を保持する配列。
     * 自信が根である要素は負の数を持つ。
     * （負の数の場合、絶対値がその集合の要素数）
     */
    public int[] upper;

    /**
     * コンストラクタ。
     * @param n 対象となる要素数
     */
    public DJSet(int n) {
        upper = new int[n];
        Arrays.fill(upper, -1);
    }

    /**
     * 要素xが含まれる集合のrootを求める。
     * rootを求める過程で経路圧縮しrootが親になるようにする。
     * @param x 要素x
     * @return 要素xが含まれる集合のroot
     */
    public int root(int x) {
        return upper[x] < 0 ? x : (upper[x] = root(upper[x]));
    }

    /**
     * 要素xと要素yが同じ集合か判定する。
     * @param x 要素x
     * @param y 要素y
     * @return 要素xと要素yが同じ集合であればtrue、そうでなければfalseを返す。
     */
    public boolean same(int x, int y) {
        return root(x) == root(y);
    }

    /**
     * 要素xと要素yを含むそれぞれの集合を合併する。
     * 要素数の多い集合のrootを残し、少ないほうのrootをその下に合併する。
     * @param x 要素x
     * @param y 要素y
     * @return 要素xと要素yを含むそれぞれの集合を合併できたらtrue、もともと同じ集合で合併しなかったらfalseを返す。
     */
    public boolean union(int x, int y) {
        x = root(x);
        y = root(y);
        if (x != y) {
            if (upper[y] < upper[x]) {
                int t = x;
                x = y;
                y = t;
            }
            upper[x] += upper[y];
            upper[y] = x;
        }
        return x != y;
    }

    /**
     * 集合の数を数える。
     * @return 集合の数
     */
    public int countSet() {
        int c = 0;
        for (int u : upper) {
            if (u < 0)
                c++;
        }
        return c;
    }

    /**
     * 要素ｘが含まれる集合の要素数を数える。
     * @param x 要素ｘ
     * @return 要素ｘが含まれる集合の要素数
     */
    public int countElement(int x) {
        return upper[root(x)] * -1;
    }
}