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 int long long
typedef long long ll;
using namespace std;

struct edge{
    int to, cost;
};

struct tr{
    int u,v,cost;
};

ll n,m;
vector<vector<edge>> g(1e5+10);

signed main(){
    cin >> n >> m;
    vector<bool> used1(n+10,false);
    vector<bool> used(n+10,false);
    ll f1,f2,f3;
    bool ans = true;
    vector<tr> zaps;
    for(int i=0;i<m;++i){
        cin >> f1 >> f2 >> f3;
        if(used1[f1] && used1[f2]){
            zaps.push_back({f1,f2,f3});
            continue;
        }
        used1[f1] = true;
        used1[f2] = true;
        g[f1].push_back({f2,f3});
        g[f2].push_back({f1,-f3});
    }

    vector<int> d(n+1, 1e15);

    for(int i=1;i<n;++i){
        if(!used[i]){
            int s = i;

            d[s] = 0;
            priority_queue<pair<int,int>> q;
            q.push({0,s});

            while (!q.empty()){
                int v = q.top().second;
                //cout << "v" << v << "\n";
                used[v] = true;
                int w1 = -(q.top().first);
                q.pop();
                if (w1 <= d[v]){
                    for (int j=0;j<g[v].size();++j){
                        int to = g[v][j].to;
                        int w = g[v][j].cost;
                        if (d[v]+w<d[to]){
                            d[to] = d[v]+w;
                            q.push({-d[to],to});
                        }
                    }
                }
            }
        }
        //cout << i << "\n";
    }


    for(int i=0;i<zaps.size();++i){
        if(d[zaps[i].v]-d[zaps[i].u] != zaps[i].cost){
            ans = false;
            break;
        }
    }

    if(ans){
        cout << "Yes";
    }
    else{
        cout << "No";
    }
}