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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(void) {
  int n, m, i, j, k;
  scanf("%d%d", &n, &m);
  int l[m], r[m], d[m];
  int x[n + 1], sum[n + 1];
  char f1[n + 1];
  for(i = 0; i <= n; ++i) f1[i] = 0, sum[i] = 0;
  for(i = 0; i < m; ++i) {
    scanf("%d%d%d", &l[i], &r[i], &d[i]);
    sum[l[i]]++;
    sum[r[i]]++;
  }
  int num = 0, numn = 0;
  for(i = 0; i <= n; ++i) if(num < sum[i]) numn = i, num = sum[i];
  f1[numn] = 1;
  x[numn] = 0;
  num = 0;
  for(i = 0; i < m; ++i) {
    if(r[i] || l[i] == numn) {
      int t;
      t = l[i], l[i] = l[num], l[num] = t;
      t = d[i], d[i] = d[num], d[num] = t;
      t = r[i], r[i] = r[num], r[num] = t;
      ++num;
    }
  }
  int tt = 1;
  while( 1 ) {
    int temp = 1, mm = 0;
    if(tt) mm = 1;
    tt = (tt + 1) % 2;
    for(i = 0; i < m; ++i) {
      if(f1[l[i]] && f1[r[i]]) {
        if(x[r[i]] - x[l[i]] != d[i]) {
          printf("No");
          return 0;
        }
        tt = 0;
      } else if(f1[l[i]] && !f1[r[i]]) {
        f1[r[i]] = 1;
        x[r[i]] = x[l[i]] + d[i];
        tt = 0;
      } else if(!f1[l[i]] && f1[r[i]]) {
        f1[l[i]] = 1;
        x[l[i]] = x[r[i]] - d[i]; 
        tt = 0;
      } else {
        temp = 0;
        if(mm) {
          mm = 0;
          x[l[i]] = 0;
          f1[l[i]] = 1;
          x[r[i]] = d[i];
          f1[r[i]] = 1;
        }
      }
    }
    if(temp) break;
  }
  printf("Yes");
  return 0;
}