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 <iostream>
#include <vector>
#include <array>
#include <set>
#include <cmath>
#include <string>
#include <sstream>
#include <queue> 
#include <list>
#include <stack>
#include <unordered_map>
#include <algorithm> 
#include <numeric>  
#include <functional>
#include <cassert>
#include <bitset>
#include <cctype>  
#include <iomanip> 
#include <limits>  
#include <regex>
// std::numeric_limits<int>::max(), min()
// std::setprecision(X)
// std::islower(), std::isupper(), std::tolower, std::toupper
// std::accumulate(m.begin(), m.end(), 0LL);
// std::sort(begin(),end()) // 0, 1, 2, 3 ,4 (default, less);
// std::priority_queue<int, vector<int>, less<int>> que; // 4, 3, 2, 1 (defualt, less) : 
// if ( std::regex_match(S, std::regex("(dream|dreamer|erase|eraser)*")) ) std::cout << "YES" << std::endl;
 
#define rep(i, n) for(int i = 0; i < (int)(n);    i++)
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
typedef long long ll;

ll gcd ( ll a, ll b ) { if ( b == 0 ) return a; else return gcd ( b, a%b ); }
ll lcm ( ll a, ll b ) { return a * b / gcd(a, b); }


template <class T> class UnionFind {

  private:
    struct subset  { 
      T parent; 
      T rank;
      T diff_weight;
      T sizes;
    }; 
    std::vector<subset> subsets;

  public:
    UnionFind (T N) {
      subsets.reserve(N);
      for(T i=0; i<N; i++){
        subsets.push_back( { i, 0, 0, 1 } );
      }
    }

    T Find (T x) { 
      if (subsets[x].parent == x) return x;
      subsets[x].diff_weight += subsets[subsets[x].parent].diff_weight;
      return subsets[x].parent = Find(subsets[x].parent); 
    } 

    T weight(T x) {
      Find(x);
      return subsets[x].diff_weight;
    }

    bool Same (T x, T y) {
        return Find(x) == Find(y);
    }

    bool Union (T x, T y, T w) {
      w += weight(x); w -= weight(y);

      T xroot = Find(x); 
      T yroot = Find(y); 
      if ( xroot == yroot ) return false;

      if (subsets[xroot].rank < subsets[yroot].rank) {
        subsets[xroot].parent = yroot;
        w = -w;
        subsets[yroot].sizes += subsets[xroot].sizes;
      }
      else if (subsets[xroot].rank > subsets[yroot].rank) {
        subsets[yroot].parent = xroot;
        subsets[xroot].sizes += subsets[yroot].sizes;
      }
      else { 
        subsets[yroot].parent = xroot; 
        subsets[xroot].rank++; 
        subsets[xroot].sizes += subsets[yroot].sizes;
      }
      subsets[yroot].diff_weight = w;
      return true;
    }

    T diff(T x, T y) {
      return weight(y) - weight(x);
    }

    T size (T x) {
      return subsets[Find(x)].sizes;
    }
};

int main (void) {
  std::cin.tie(0);  std::ios::sync_with_stdio(false);
  int N, M; std::cin >> N >> M;
  UnionFind<int> uf(N);
  rep(i,M) {
    int l,r,d; std::cin >> l >> r >> d; l--, r--;
    if ( uf.Same(l,r) ) {
      if ( uf.diff(l,r) != d ) {
        std::cout << "No" << std::endl;
        return 0;
      }
    }
    else uf.Union(l,r,d);
  }
  std::cout << "Yes" << std::endl;
  return 0;
}