human
What is a Python code to determine if there exists a set of values (x_1, x_2, given the following:
I found an interesting problem on https://atcoder.jp/contests/abc087/tasks/arc090_b:
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-----
3 3
1 2 1
2 3 1
1 3 2

-----Sample Output-----
Yes

Some possible sets of values (x_1, x_2, x_3) are (0, 1, 2) and (101, 102, 103).

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is Python code to determine if there exists a set of values (x_1, x_2, .

import sys
sys.setrecursionlimit(10**9)

read=sys.stdin.read
class Unionfind():
    def __init__(self,n):
        self.parents=[-1]*n
        self.dist=[0]*n 
    
    def find(self,x):
        if self.parents[x]<0:
            return x,self.dist[x]
        else:
            tmp=self.find(self.parents[x])
            self.parents[x]=tmp[0]
            self.dist[x]+=tmp[1]
            return self.parents[x],self.dist[x]
    
    def union(self,x,y,d):
        rx=self.find(x)[0]
        ry=self.find(y)[0]
        diff=self.dist[y]-self.dist[x]-d
        if rx==ry:
            if diff!=0:
                return True
            return False
        if diff<0:
            rx,ry=ry,rx
            diff=-diff
        self.parents[ry]=min(self.parents[ry],self.parents[rx]-diff)
        self.parents[rx]=ry
        self.dist[rx]=diff
        return False
    
def main():
    n,m,*lrd=list(map(int,read().split()))
    v=Unionfind(n)
    for l,r,d in zip(*[iter(lrd)]*3):
        if v.union(l-1,r-1,d):
            print('No')
            break
    else:
        print('Yes')
def __starting_point():
    main()

__starting_point()