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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

#define INF 100100100
#define MOD 1000000007

using namespace std;
using ll = long long;
using ull = unsigned long long ;

int main(int argc, char const* argv[]){
   ios::sync_with_stdio(false);
   cin.tie(0);
   ull N,M;
   cin>>N>>M;
   vector<ll>x(N,0);
   vector<bool>used(N,false);
   for (ull i = 0; i < M; i++) {
      ull L,R,D;
      cin>>L>>R>>D;
      L--,R--;
      if(used[R]&&used[L]){
         if(x[R]-x[L]!=D){
            cout<<"No"<<endl;
            return 0;
         }
      }
      else if(used[R]){
         x[L]=x[R]-D;
         used[L]=true;
      }
      else {
         x[R]=x[L]+D;
         used[R]=true;
         used[L]=true;
      }
   }
   if(*max_element(x.begin(),x.end()) > 1000000000 
         || -*min_element(x.begin(),x.end()) > 1000000000){
      cout<<"No"<<endl;
      return 0;
   }

   ull dis = *max_element(x.begin(),x.end()) - *min_element(x.begin(),x.end());
   if(dis > 1000000000)cout<<"No"<<endl;
   else cout<<"Yes"<<endl;
   return 0;
}