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 <algorithm>
#include <string.h>
#include <math.h>
#include <queue>

using namespace std;

#define pb push_back
#define mp make_pair
#define ll long long
#define ull unsigned ll
#define db double
#define INF 0x3f3f3f3f
#define MOD 1000000007
#define PII pair<int, int>

const int N=100010,M=200010;
struct edge {
    int to,next,val;
} e[M];
int head[N],tot=0;
void add_edge(int u,int v,int val) {
    e[tot].to=v;
    e[tot].val=val;
    e[tot].next=head[u];
    head[u]=tot++;
}
int in[N],dis[N];

int n,m,l,r,d;
int main()
{
    scanf("%d%d",&n,&m);
    memset(head,-1,sizeof(head));
    memset(dis,INF,sizeof(dis));
    for (int i=1;i<=m;i++) {
        scanf("%d%d%d",&l,&r,&d);
        add_edge(l,r,d);
        in[r]++;
    }
    queue<PII> q;
    for (int i=1;i<=n;i++) {
        if (!in[i]) q.push(mp(i,0));
    }
    if (m>0&&q.empty()) {
        printf("NO\n");
        return 0;
    }
    int cnt=0;
    while (!q.empty()) {
        PII cur=q.front();//printf("cur:%d dis:%d\n",cur.first,cur.second);
        q.pop();
        if (dis[cur.first]==INF) {
            cnt++;
            dis[cur.first]=cur.second;
            for (int i=head[cur.first];i!=-1;i=e[i].next) {
                int to=e[i].to;
                q.push({to,cur.second+e[i].val});
            }
        } else {
            if (dis[cur.first]!=cur.second) {
                printf("NO\n");
                return 0;
            }
        }
    }
    if (cnt==n) printf("YES\n");
    else printf("NO\n");
}