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<cstdio>
#include<cstring>
#include<iostream>
#include<string>
#include<map>
#include<vector>
#include<queue>
#include<set>
#include<stack>
#include<sstream>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=100000+10;
int sz[N],n,x,m,vis[N],a,b;
struct {
    int s,st,en,t,d;
}p[N];
struct P{
    int l,r,d;
    friend bool operator<(const P &p1,const P &p2){
        if(vis[p1.l]||vis[p1.r]) return 0;
        if(vis[p2.l]||vis[p2.r]) return 1;
        return 1;
    }
};
int find(int now){
    if(p[now].st!=now) return p[now].st;
    return now;
}
int main() {
    while(~scanf("%d%d",&n,&m)){
        bool flag=1;
        memset(sz,0,sizeof(sz));
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=n;i++){
            p[i].s=-1;
            p[i].t=-1;
            p[i].st=i;
            p[i].en=-1;
            p[i].d=0;
        }
        priority_queue<P>q;
        P pp;
        for(int i=0;i<m;i++){
            scanf("%d%d%d",&pp.l,&pp.r,&pp.d);
            q.push(pp);

        }
        while(!q.empty()){
            pp=q.top();
            q.pop();
            int l=pp.l,r=pp.r,d=pp.d;
            //printf("%d %d %d\n",l,r,d);
            if(!flag) continue;
            a=find(l);
            b=find(r);
            if(a==b){
                if(sz[l]+d+p[p[l].st].d!=sz[r]+p[p[r].st].d) flag=0;
            }
            else {
                if(vis[l]&&!vis[r]){
                    sz[r]=sz[l]+d;
                    p[p[a].en].t=r;
                    p[r].s=p[a].en;
                    p[a].en=r;
                    p[r].st=a;
                }
                else if(!vis[l]&&vis[r]){
                    sz[l]=sz[r]-d;
                    p[p[b].en].t=l;
                    p[l].s=p[b].en;
                    p[b].en=l;
                    p[l].st=b;

                }
                else if(!vis[l]&&!vis[r]){
                    sz[r]=sz[l]+d;
                    p[p[a].en].t=r;
                    p[r].s=p[a].en;
                    p[a].en=r;
                    p[r].st=a;
                }
                else {
                    int xxx=sz[l]+d-sz[r];
                    p[b].d+=xxx;
                    p[p[a].en].t=p[b].st;
                    p[b].st=a;
                    p[b].s=p[a].en;
                    p[a].en=p[b].en;
                }

            }
            vis[pp.l]=1;
            vis[pp.r]=1;

        }
        if(flag) printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}