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(){
  int n,m,big,i,f=0,count=0;
  scanf("%d %d",&n,&m);
  int L[200000],R[200000],D[200000];
  for(i=0;i<m;i++){
    scanf("%d %d %d",&L[i],&R[i],&D[i]);
  }
  int x[200000],y[200000],z[200000];
  for(i=0;i<m;i++){
    x[i]=0;
    y[i]=0;
    z[i]=0;
  }
  for(big=0;big>=0;big++){
    for(i=0;i<m;i++){
      if(i==0){
        x[L[i]]=1;
        x[R[i]]=1;
        y[L[i]]=0;
        y[R[i]]=D[i];
        count += 1;
      }else{
        if(x[L[i]]==1 && x[R[i]]==1){
          if(y[R[i]]-y[L[i]]!=D[i]){
            f = 1;
            break;
          }
        }else if(x[L[i]]==0 && x[R[i]]==0){
          if(big==3){
            y[L[i]]=0;
            y[R[i]]=D[i];
          }          
        }else if(x[L[i]]==0 && x[R[i]]==1){
          y[L[i]]=y[R[i]]-D[i];
          x[L[i]]=1;
          z[i]=1;
          count += 1;
        }else{
          y[R[i]]=y[L[i]]+D[i];
          x[R[i]]=1;
          z[i]=1;
          count += 1;
        }
      }
    }
      if(count==m-1 && big ==0){
      if(f==1){
        printf("No\n");
        big = -1;
        break;
      }else{
        printf("Yes\n");
        big = -1;
        break;
      }
    }
    if(f!=0){
      printf("No\n");
    }
    if(big == 20 && f==0){
      printf("Yes\n");
      break;
    }
  }
  return 0;
}