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 9.2.1) (judge verdict: Wrong Answer)

#include <algorithm>
#include <cassert>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <vector>
#include <limits.h>

using namespace std;

typedef long long ll;

template<class T>
inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}

const int INF = INT_MAX - 10000;

// graphs[i] means {to, dist} from vertex i
vector<vector<pair<int, int>>> graph;

vector<int> deps;

vector<int> checked;

bool bfs(const int start, vector<int> &dists, const int round) {
    queue<int> que;
    dists[start] = 0;
    que.push(start);
    checked[start] = round;

    while (!que.empty()) {
        auto cur = que.front();
        // printf("cur: %d\n", cur);
        que.pop();
        for (auto p : graph[cur]) {
            int next_v = p.first;
            int next_dist = dists[cur] + p.second;
            // deps[next_v]--;
            if (checked[next_v] > 0 && checked[next_v] != checked[cur]) {
                // do nothing
            } else if (dists[next_v] == INF) {
                dists[next_v] = next_dist;
                que.push(next_v);
                checked[next_v] = round;
            } else if (dists[next_v] == next_dist) {
                // do nothing
            } else {
                // printf("cur: %d, next_v: %d, dists[next_v]: %d, next_dist: %d\n", cur, next_v, dists[next_v], next_dist);
                return false;
            }
        }
    }

    return true;
}

int main(void) {
    int N, M;
    cin >> N >> M;

    graph.resize(N);
    deps.resize(N, 0);
    for (int i = 0; i < M; i++) {
        int l, r, d;
        cin >> l >> r >> d;
        l--;
        r--;
        graph[r].push_back(make_pair(l, d));
        deps[l]++;
    }

    checked.resize(N, false);

    bool ok = true;
    int round = 1;
    for (int i = 0; i < N; i++) {
        if (deps[i] == 0) {
            vector<int> dists(N, INF);
            if (!bfs(i, dists, round)) {
                ok = false;
                break;
            }
            round++;
        }
    }

    for (int i = 0; i < N; i++) {
        if (checked[i] == 0) {
        // if (deps[i] > 0) {
            ok = false;
        }
    }

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

    return 0;
}