stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include <bits/stdc++.h>

#define LL long long

#define mp make_pair

#define pb push_back

#define INF (1<<29)

using namespace std;

typedef pair<int,int>pii;

int n,m;

vector<pii>G[200010];

int col[200010];

int vis[200010];

bool dfs(int x,int co){

//	cout<<"In:"<<x<<' '<<co<<endl;

	if(vis[x]!=-1)return col[x]==co;

	vis[x]=1;

	col[x]=co;

	for(pii i:G[x]){

		if(dfs(i.first,co+i.second)==0)return false;

//		else{

//			if(vis[i.first]!=co+i.second)return false;

//		}

	}

	return true;

}

int main(){

//	freopen("produce/People.txt","r",stdin);

	cin>>n>>m;

	memset(vis,-1,sizeof(vis));

	if(m==0){

		cout<<"Yes\n";

		return 0;

	}

	for(int i=1;i<=m;i++){

		int l,r,d;

		cin>>l>>r>>d;

		G[l].pb(mp(r,d));

		G[r].pb(mp(l,-d));

	}

	for(int i=1;i<=n;i++){

		if(vis[i]!=-1)continue;

		memset(col,0,sizeof(col));

		if(dfs(i,0)==0){

			cout<<"No\n";

			return 0;

		}

//		cout<<"-----------------------\n";

	}

	cout<<"Yes\n";

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 8.2623016525844x):
#include <bits/stdc++.h>

#define INL inline

#define REG register

#define DB double

#define LDB long double

#define ULL unsigned long long

#define LL long long



#define RPT(i,x,y) for (REG int i=(x);i<(y);i++)

#define DRPT(i,x,y) for (REG int i=(x);i>(y);i--)

#define MST(a,b) memset((a),(b),sizeof(a))

#define MRK() cout<<"Mark"<<endl;

#define WRT(x) cout<<#x<<" = "<<(x)<<endl;



#define MAXN 110000

#define MAXM 10000

#define MOD 998244353

#define INF 0x3f3f3f3f

#define LLINF 0x3f3f3f3f3f3f3f3f

#define EPS 1e-5



#define _ 0

using namespace std;



int pa[MAXN],dis[MAXN];



int getpa(int x)

{

	if (pa[x]==x)

		return x;

	int f=pa[x];

	pa[x]=getpa(pa[x]);

	dis[x]+=dis[f];

	return pa[x];

}



bool merge(int x,int y,int d)

{

	int px=getpa(x),py=getpa(y);

	if (px==py && d!=dis[x]-dis[y])

		return false;

	pa[px]=py;

	dis[px]=d+dis[y]-dis[x];

	return true;

}



int n,m;



int main()

{

	int x,y,z;

	cin>>n>>m;

	RPT(i,1,n+1)

		pa[i]=i;

	RPT(i,0,m)

	{

		scanf("%d%d%d",&x,&y,&z);

		if (!merge(y,x,z))

		{

			cout<<"No";

			return 0;

		}

		/*

		cout<<endl<<"--------------\n\n";

		RPT(i,1,n+1)

		{

			cout<<pa[i]<<' '<<dis[i]<<endl;

		}

		*/

	}

	cout<<"Yes";

	return ~~(0^_^0);

}
// Slower C++ solution:
#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

#define int long long

using namespace std;

int n,m,cnt=0;

int head[410000],nxt[410000],to[410000],val[410000];

int ma=0,mi=0;

int q[1000000],dis[100100],f[100100],vis[100100];

void addedge(int x,int y,int z)

{

    cnt++;

    nxt[cnt]=head[x];

    head[x]=cnt;

    to[cnt]=y;

    val[cnt]=z;

}

bool spfa(int x)

{

    int l=0,r=1;

    q[1]=x;vis[x]=1;dis[x]=0;

    while(l<r)

    {

        int u=q[++l];

        for(int i=head[u];i!=-1;i=nxt[i])

        {

            int v=to[i];

            if(dis[v]==12345678987654321)

            {

                dis[v]=dis[u]+val[i];

                ma=max(ma,dis[v]);

                mi=min(mi,dis[v]);

                if(ma-mi>1e9) return 0;

                vis[v]=1;

                if(!f[v]) q[++r]=v,f[v]=1;

            }

            else if(dis[v]!=dis[u]+val[i]) return 0;

        }

        f[u]=0;

    }

    return 1;

}

signed main()

{

//	freopen("line.in","r",stdin);

//	freopen("line.out","w",stdout);

    memset(head,-1,sizeof(head));

    for(int i=1;i<=100000;i++)

        dis[i]=12345678987654321;

    scanf("%lld%lld",&n,&m);

    for(int i=1;i<=m;i++)

    {

        int x,y,z;

        scanf("%lld%lld%lld",&x,&y,&z);

        addedge(x,y,z);

        addedge(y,x,-z);

    }

    for(int i=1;i<=n;i++)

        if(!vis[i]) {if(!spfa(i)) {printf("No");return 0;}}

    printf("Yes");

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.115844582714549x):
#include <bits/stdc++.h>

#define INL inline

#define REG register

#define DB double

#define LDB long double

#define ULL unsigned long long

#define LL long long



#define RPT(i,x,y) for (REG int i=(x);i<(y);i++)

#define DRPT(i,x,y) for (REG int i=(x);i>(y);i--)

#define MST(a,b) memset((a),(b),sizeof(a))

#define MRK() cout<<"Mark"<<endl;

#define WRT(x) cout<<#x<<" = "<<(x)<<endl;



#define MAXN 110000

#define MAXM 10000

#define MOD 998244353

#define INF 0x3f3f3f3f

#define LLINF 0x3f3f3f3f3f3f3f3f

#define EPS 1e-5



#define _ 0

using namespace std;



int pa[MAXN],dis[MAXN];



int getpa(int x)

{

	if (pa[x]==x)

		return x;

	int f=pa[x];

	pa[x]=getpa(pa[x]);

	dis[x]+=dis[f];

	return pa[x];

}



bool merge(int x,int y,int d)

{

	int px=getpa(x),py=getpa(y);

	if (px==py && d!=dis[x]-dis[y])

		return false;

	pa[px]=py;

	dis[px]=d+dis[y]-dis[x];

	return true;

}



int n,m;



int main()

{

	int x,y,z;

	cin>>n>>m;

	RPT(i,1,n+1)

		pa[i]=i;

	RPT(i,0,m)

	{

		scanf("%d%d%d",&x,&y,&z);

		if (!merge(y,x,z))

		{

			cout<<"No";

			return 0;

		}

		/*

		cout<<endl<<"--------------\n\n";

		RPT(i,1,n+1)

		{

			cout<<pa[i]<<' '<<dis[i]<<endl;

		}

		*/

	}

	cout<<"Yes";

	return ~~(0^_^0);

}
// Slower C++ solution:
#include<bits/stdc++.h>

#define R(i,j,k) for(int i=j;i<=k;i++)

using namespace std;

const int maxn=1e5+5;

int n,m;

struct Union_Set {

	int p[maxn],v[maxn];

	Union_Set() {

		R(i,1,maxn-1) p[i]=i;

	}

	void fp(int x) {

		if(x==p[x]) return;

		fp(p[x]);

		v[x]+=v[p[x]];

		p[x]=p[p[x]];

	}

	bool merge(int x,int y,int val) {

		fp(x); fp(y);

		int X=p[x],Y=p[y];

		if(X==Y) {

			int v1=-v[x],v2=-v[y];

			if(v1+val==v2) return true;

			return false;

		}

		val=val+v[y]-v[x];

		p[X]=Y;

		v[X]=val;

		return true;

	}

}us;

int main() {

#ifdef LOCAL

	//freopen("line.in","r",stdin);

#endif

	cin>>n>>m;

	R(i,1,m) {

		int Li,Ri,Di;

		cin>>Li>>Ri>>Di;

		if(!us.merge(Li,Ri,Di)) {

			cout<<"No"<<endl;

			return 0;

		}

	}

	cout<<"Yes"<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4029005066934876x):
#include <bits/stdc++.h>

#define INL inline

#define REG register

#define DB double

#define LDB long double

#define ULL unsigned long long

#define LL long long



#define RPT(i,x,y) for (REG int i=(x);i<(y);i++)

#define DRPT(i,x,y) for (REG int i=(x);i>(y);i--)

#define MST(a,b) memset((a),(b),sizeof(a))

#define MRK() cout<<"Mark"<<endl;

#define WRT(x) cout<<#x<<" = "<<(x)<<endl;



#define MAXN 110000

#define MAXM 10000

#define MOD 998244353

#define INF 0x3f3f3f3f

#define LLINF 0x3f3f3f3f3f3f3f3f

#define EPS 1e-5



#define _ 0

using namespace std;



int pa[MAXN],dis[MAXN];



int getpa(int x)

{

	if (pa[x]==x)

		return x;

	int f=pa[x];

	pa[x]=getpa(pa[x]);

	dis[x]+=dis[f];

	return pa[x];

}



bool merge(int x,int y,int d)

{

	int px=getpa(x),py=getpa(y);

	if (px==py && d!=dis[x]-dis[y])

		return false;

	pa[px]=py;

	dis[px]=d+dis[y]-dis[x];

	return true;

}



int n,m;



int main()

{

	int x,y,z;

	cin>>n>>m;

	RPT(i,1,n+1)

		pa[i]=i;

	RPT(i,0,m)

	{

		scanf("%d%d%d",&x,&y,&z);

		if (!merge(y,x,z))

		{

			cout<<"No";

			return 0;

		}

		/*

		cout<<endl<<"--------------\n\n";

		RPT(i,1,n+1)

		{

			cout<<pa[i]<<' '<<dis[i]<<endl;

		}

		*/

	}

	cout<<"Yes";

	return ~~(0^_^0);

}
// Slower C++ solution:
# include <cstring>

# include <cstdio>

# include <iostream>

# include <queue>

# define N 100050

# define M 200050 

# define WQD 0x7f7f7f7f

int n;

int m;

struct Edge{int next,to, w;}  e[M<<1];

int head[N+1]; 

int dis[N+1];

bool vis[N+1];

bool book[ N + 1 ] ;

bool spfa(int s);

using namespace std;

int main()

{

	memset(dis,0x7f,sizeof(dis));

	memset(book,0,sizeof(book));

	ios::sync_with_stdio ( false ) ;

	int u,v,w;

	memset(head,-1,sizeof(head));

	cin>>n>>m;

	for(int i=1; i<=m; i++)

	{

		cin>>u>>v>>w;

		e[i]=(Edge){head[u],v,w};head[u]=i;

		e[m+i]=(Edge){head[v],u,-w};head[v]=m+i;

	}

	for(int i=1; i<=n; i++)

		if(!vis[i]) 

		{

			vis[i]=1;

			if(!spfa(i))  {cout<<"No";return 0;}

		}

	cout<<"Yes";

	return 0;

}

bool spfa(int s)

{

 	queue<int>q;

//	memset(book,0,sizeof(book));

//	memset(dis,0x7f,sizeof(dis));

	dis[s]=0;

	q.push(s);book[s]=1;

	while(q.size()>0)

	{

		int k=q.front();q.pop();

		book[k]=0;

		for(int i=head[k]; i!=-1; i=e[i].next)

		{

			int v=e[i].to;

			if(dis[v]==WQD ) 

			{				

				dis[v]=dis[k]+e[i].w;

				if( ! book[v])	{q.push(v);	book[v]=1;}

				vis[v]=1;

			}

			else if(  (dis[v]!=dis[k]+e[i].w) )	return false;

		}		

    }

	return true;

}

// Faster C++ solution for the same problem (measured speedup: 10.350351924722242x):
// luogu-judger-enable-o2

// SeptEtavioxy

#include<cstdio>

#include<cctype>

#include<cstring>

#include<algorithm>

#include<cmath>

#define re register

#define ll long long

#define il inline

#define rep(i,s,t) for(re int i=(s);i<=(t);i++)

#define rev_rep(i,s,t) for(re int i=(s);i>=(t);i--)

#define each(i,u) for(int i=head[u];i;i=bow[i].nxt)

#define pt(ch) putchar(ch)

#define pti(x) printf("%d",x)

#define ptll(x) printf("%lld",x)

#define file(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)

using namespace std;

// c_ints

il int ci(){

	re char ch;int f=1;

	while(!isdigit(ch=getchar()))f= ch=='-'?-1:1 ;

	re int x= ch^'0';

	while(isdigit(ch=getchar()))x=(x*10)+(ch^'0');

	return f*x;

}

enum{N=100024};

int fa[N],dis[N];

void fd(int x){

	if( fa[x]==x ) return ;

	fd(fa[x]);

	dis[x]+= dis[fa[x]];

	fa[x]= fa[fa[x]];

}

int main(){

	int n=ci(), m=ci();

	rep(i,1,n) fa[i]= i;

	while( m-- ){

		int l=ci(), r=ci(), d=ci();

		fd(l), fd(r);

		if( fa[l]==fa[r] ){

			if( dis[r]-dis[l]!=d ){

				puts("No");

				return 0;

			}

		}

		else{

			int fl=fa[l], fr=fa[r];

			int x= dis[l]+d-dis[r];

//			printf("%d %d\n",fl,fr);

//			if( x<0 ){

//				puts("No");

//				return 0;

//			}

			dis[fr]= x;

			fa[fr]= fl;

		}

//		printf("fa :");rep(i,1,n) printf("%d ",fa[i]); pt('\n');

//		printf("dis:");rep(i,1,n) printf("%d ",dis[i]); pt('\n');

	}

	puts("Yes");

	return 0;

}
// Slower C++ solution:
# include <cstring>

# include <cstdio>

# include <iostream>

# include <queue>

# define N 100050 //N为顶点个数 

# define M 200050 //N为顶点个数 

# define WQD 0x7f7f7f7f//无穷大 

int n;//图的顶点个数

int m;//图的边数 

struct Edge{int next,to, w;}  e[M<<1];//不用0号单元 

int head[N+1]; 

int dis[N+1];//存放源点到其他点的路径

bool vis[N+1];

bool book[N+1];

bool spfa(int s);//差分约束判断 

using namespace std;

int main()

{

	ios::sync_with_stdio ( false ) ;

	int u,v,w;

	memset(head,-1,sizeof(head));

	cin>>n>>m;

	for(int i=1; i<=m; i++)  //本题约束关系为=，要添加两条边 

	{

		cin>>u>>v>>w;//输入第i条边 ,i为编号 

		e[i]=(Edge){head[u],v,w};head[u]=i;

		e[m+i]=(Edge){head[v],u,-w};head[v]=m+i;

	}

	memset(book,0,sizeof(book));

	memset(dis,0x7f,sizeof(dis));

	for(int i=1; i<=n; i++)

		if(!vis[i]) 

		{

			vis[i]=1;

			if(spfa(i)==false)  {cout<<"No";return 0;}

		}

	cout<<"Yes";

	return 0;

}

bool spfa(int s)

{

 	queue<int>q;

 	//memset(book,0,sizeof(book));

	//memset(dis,0x7f,sizeof(dis));

	dis[s]=0;

	q.push(s);book[s]=1;

	while(q.size()>0)//如果没松弛就松弛，如果松弛过就判断是否符合约束条件，如果自相矛盾就是不可能的数据 

	{

		int k=q.front();q.pop();//对头元素出队 

		book[k]=0;

		for(int i=head[k]; i!=-1; i=e[i].next)

		{

			int v=e[i].to;

			if(dis[v]==WQD )//如果v点没松弛过，就松弛更新dis[v]的值 

				//题目的约束条件是= 

				{				

					dis[v]=dis[k]+e[i].w;//更新 

					if(book[v]==0)	{q.push(v);	book[v]=1;}//入队标记

					vis[v]=1;//标记该点已经松弛过 

				}

			else if(dis[v]!=dis[k]+e[i].w)	return false;//这里也可以用不等于，根据题意理解差分系统约束条件 

		}		

    }

	return true;//别掉了这句 

}

// Faster C++ solution for the same problem (measured speedup: 10.337598130686542x):
// luogu-judger-enable-o2

// SeptEtavioxy

#include<cstdio>

#include<cctype>

#include<cstring>

#include<algorithm>

#include<cmath>

#define re register

#define ll long long

#define il inline

#define rep(i,s,t) for(re int i=(s);i<=(t);i++)

#define rev_rep(i,s,t) for(re int i=(s);i>=(t);i--)

#define each(i,u) for(int i=head[u];i;i=bow[i].nxt)

#define pt(ch) putchar(ch)

#define pti(x) printf("%d",x)

#define ptll(x) printf("%lld",x)

#define file(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)

using namespace std;

// c_ints

il int ci(){

	re char ch;int f=1;

	while(!isdigit(ch=getchar()))f= ch=='-'?-1:1 ;

	re int x= ch^'0';

	while(isdigit(ch=getchar()))x=(x*10)+(ch^'0');

	return f*x;

}

enum{N=100024};

int fa[N],dis[N];

void fd(int x){

	if( fa[x]==x ) return ;

	fd(fa[x]);

	dis[x]+= dis[fa[x]];

	fa[x]= fa[fa[x]];

}

int main(){

	int n=ci(), m=ci();

	rep(i,1,n) fa[i]= i;

	while( m-- ){

		int l=ci(), r=ci(), d=ci();

		fd(l), fd(r);

		if( fa[l]==fa[r] ){

			if( dis[r]-dis[l]!=d ){

				puts("No");

				return 0;

			}

		}

		else{

			int fl=fa[l], fr=fa[r];

			int x= dis[l]+d-dis[r];

//			printf("%d %d\n",fl,fr);

//			if( x<0 ){

//				puts("No");

//				return 0;

//			}

			dis[fr]= x;

			fa[fr]= fl;

		}

//		printf("fa :");rep(i,1,n) printf("%d ",fa[i]); pt('\n');

//		printf("dis:");rep(i,1,n) printf("%d ",dis[i]); pt('\n');

	}

	puts("Yes");

	return 0;

}
// Slower C++ solution:
# include <cstdio>

# include <iostream>

# include <cstring>

# include <algorithm>

# include <queue>

# define N 100001

# define M 200001

# define _cin ios::sync_with_stdio(false)

# define Inf 0x7f7f7f7f

using namespace std;

struct ege

{

	int next;

	int to;

	int w;

}e[M<<1];

int n;//人数 

int m;//信息数 

int book[N+1];

int head[N+1];

int dis[N+1];

bool vis[N+1];

bool spfa(int x);

int main()

{

	_cin;

	

	int u,v,w;

	cin>>n>>m;

	memset(head,-1,sizeof(head));

	for(int i=1;i<=m;i++)	

	{

		cin>>u>>v>>w;

		//建立双向边 

		/*e[i].next=head[u];

		e[i].to=v;

		e[i].w=w;

		head[u]=i;

		e[i+m].next=head[v];

		e[i+m].to=u;

		e[i+m].w=-w;

		head[v]=i+m;

		*/

		e[i]=(ege){head[u],v,w};head[u]=i;

		e[m+i]=(ege){head[v],u,-w};head[v]=m+i;		

	}

	memset(book,0,sizeof(book));

	memset(dis,0x7f,sizeof(dis));

	for(int i=1;i<=n;i++)

	if(!vis[i])//表示两点连通

	{

		vis[i]=1;//连通就变成真值 

		if(spfa(i)==false)	{cout<<"No"<<endl;return 0;}

	} 

	cout<<"Yes"<<endl;

	return 0;

}

bool spfa(int x)

{

	queue <int>q;

	dis[x]=0;

	q.push(x);book[x]=1;//标记已经进队列的点 

	while(q.size()>0)

	{

		int k=q.front();//取头元素

		q.pop();//头元素出队列

		book[k]=0;

		for(int i=head[k]; i!=-1; i=e[i].next)

		{

			if(dis[e[i].to]==Inf && dis[k]<Inf)

			{

				dis[e[i].to]=dis[k]+e[i].w;

				//如果下一个点没有被标记，将这个点标记，进队列 

				if(book[e[i].to]==0)	{q.push(e[i].to);book[e[i].to]=1;}

				vis[e[i].to]=1;

			} 

			//与已知矛盾，即任意两点的距离都不能有两个值 

			//表示k点已被改变值，下一个点也已经改变值，但是新值与旧值不等 

			else if(dis[k]<Inf && dis[e[i].to]<Inf && dis[e[i].to]!=dis[k]+e[i].w) return false;

		} 

	}

	return true;

}

// Faster C++ solution for the same problem (measured speedup: 11.999126390670273x):
// luogu-judger-enable-o2

// SeptEtavioxy

#include<cstdio>

#include<cctype>

#include<cstring>

#include<algorithm>

#include<cmath>

#define re register

#define ll long long

#define il inline

#define rep(i,s,t) for(re int i=(s);i<=(t);i++)

#define rev_rep(i,s,t) for(re int i=(s);i>=(t);i--)

#define each(i,u) for(int i=head[u];i;i=bow[i].nxt)

#define pt(ch) putchar(ch)

#define pti(x) printf("%d",x)

#define ptll(x) printf("%lld",x)

#define file(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)

using namespace std;

// c_ints

il int ci(){

	re char ch;int f=1;

	while(!isdigit(ch=getchar()))f= ch=='-'?-1:1 ;

	re int x= ch^'0';

	while(isdigit(ch=getchar()))x=(x*10)+(ch^'0');

	return f*x;

}

enum{N=100024};

int fa[N],dis[N];

void fd(int x){

	if( fa[x]==x ) return ;

	fd(fa[x]);

	dis[x]+= dis[fa[x]];

	fa[x]= fa[fa[x]];

}

int main(){

	int n=ci(), m=ci();

	rep(i,1,n) fa[i]= i;

	while( m-- ){

		int l=ci(), r=ci(), d=ci();

		fd(l), fd(r);

		if( fa[l]==fa[r] ){

			if( dis[r]-dis[l]!=d ){

				puts("No");

				return 0;

			}

		}

		else{

			int fl=fa[l], fr=fa[r];

			int x= dis[l]+d-dis[r];

//			printf("%d %d\n",fl,fr);

//			if( x<0 ){

//				puts("No");

//				return 0;

//			}

			dis[fr]= x;

			fa[fr]= fl;

		}

//		printf("fa :");rep(i,1,n) printf("%d ",fa[i]); pt('\n');

//		printf("dis:");rep(i,1,n) printf("%d ",dis[i]); pt('\n');

	}

	puts("Yes");

	return 0;

}
// Slower C++ solution:
#include <cstdio>

#include <iostream>

#include <algorithm>

#include <cstring>

#include <queue>

#define N 100001

#define M 200001

#define INF 0x7fffffff

#define ri register int

#define in inline

using namespace std;

struct edge

{

	int to,w,next;

}e[M<<1];

//vis数组表示已松弛的点

//book数组表示进入队列的点 

int head[N],vis[N],dis[N];

int m,n;

bool book[N];

in bool spfa(int s);

int main()

{

	ios::sync_with_stdio(false);

	int u,v,w;

	cin>>n>>m;

	for(ri i=1;i<=m;i++)

	{

		cin>>u>>v>>w;

		e[i]=(edge) {v,w,head[u]}; head[u]=i;

		e[i+m]=(edge) {u,-w,head[v]}; head[v]=i+m;

	}

	//一组数据为一个问题，不能将fill数组放在spfa里 

	fill(dis+1,dis+n+1,INF);

	//所有约束条件需要考虑到

	//可能有多个分块，前10个约束条件与后10个 无关联 

	for(ri i=1;i<=n;i++)

		if(!vis[i])

		{

			if(spfa(i))

			{cout<<"No"; return 0;}

			vis[i]=1;

		}	

	cout<<"Yes";

	return 0;

}

in bool spfa(int s)

{

	queue <int> q;

	int k;

	q.push(s); book[s]=1;

	dis[s]=0;

	while(q.size())

	{

		k=q.front(); q.pop();

		book[k]=0;//出队列应置为0 

		for(ri i=head[k]; i ;i=e[i].next)

		{

			int v=e[i].to;

			if(!vis[v]&& dis[k]<INF)

			{

				dis[v]=dis[k]+e[i].w;

				if(!book[v])

				{q.push(v); book[v]=1;}

				vis[v]=1;

			}

			else if(dis[k]<INF && dis[v]<INF && dis[v]!=dis[k]+e[i].w) return true;

			//此时相当于自相矛盾

			//思想：即先认定一开始所给条件均正确（此时松弛）

		    //最后再来验证接下来的条件是否满足原先条件 

		}

	}

	return false;

}

// Faster C++ solution for the same problem (measured speedup: 5.4912300750161265x):
// luogu-judger-enable-o2

// SeptEtavioxy

#include<cstdio>

#include<cctype>

#include<cstring>

#include<algorithm>

#include<cmath>

#define re register

#define ll long long

#define il inline

#define rep(i,s,t) for(re int i=(s);i<=(t);i++)

#define rev_rep(i,s,t) for(re int i=(s);i>=(t);i--)

#define each(i,u) for(int i=head[u];i;i=bow[i].nxt)

#define pt(ch) putchar(ch)

#define pti(x) printf("%d",x)

#define ptll(x) printf("%lld",x)

#define file(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)

using namespace std;

// c_ints

il int ci(){

	re char ch;int f=1;

	while(!isdigit(ch=getchar()))f= ch=='-'?-1:1 ;

	re int x= ch^'0';

	while(isdigit(ch=getchar()))x=(x*10)+(ch^'0');

	return f*x;

}

enum{N=100024};

int fa[N],dis[N];

void fd(int x){

	if( fa[x]==x ) return ;

	fd(fa[x]);

	dis[x]+= dis[fa[x]];

	fa[x]= fa[fa[x]];

}

int main(){

	int n=ci(), m=ci();

	rep(i,1,n) fa[i]= i;

	while( m-- ){

		int l=ci(), r=ci(), d=ci();

		fd(l), fd(r);

		if( fa[l]==fa[r] ){

			if( dis[r]-dis[l]!=d ){

				puts("No");

				return 0;

			}

		}

		else{

			int fl=fa[l], fr=fa[r];

			int x= dis[l]+d-dis[r];

//			printf("%d %d\n",fl,fr);

//			if( x<0 ){

//				puts("No");

//				return 0;

//			}

			dis[fr]= x;

			fa[fr]= fl;

		}

//		printf("fa :");rep(i,1,n) printf("%d ",fa[i]); pt('\n');

//		printf("dis:");rep(i,1,n) printf("%d ",dis[i]); pt('\n');

	}

	puts("Yes");

	return 0;

}
// Slower C++ solution:
# include <cstring>

# include <cstdio>

# include <iostream>

# include <queue>

# define N 100050 //N为顶点个数 

# define M 200050 //N为顶点个数 

# define WQD 0x7f7f7f7f//无穷大 

int n;//图的顶点个数

int m;//图的边数 

struct Edge{int next,to, w;}  e[M<<1];//不用0号单元 

int head[N+1]; 

int dis[N+1];//存放源点到其他点的路径

bool vis[N+1];

bool book[N+1];

bool spfa(int s);//差分约束判断 

using namespace std;

int main()

{

	ios::sync_with_stdio ( false ) ;

	int u,v,w;

	memset(head,-1,sizeof(head));

	cin>>n>>m;

	for(int i=1; i<=m; i++)  //本题约束关系为=，要添加两条边 

	{

		cin>>u>>v>>w;//输入第i条边 ,i为编号 

		e[i]=(Edge){head[u],v,w};head[u]=i;

		e[m+i]=(Edge){head[v],u,-w};head[v]=m+i;

	}

	memset(book,0,sizeof(book));

	memset(dis,0x7f,sizeof(dis));

	for(int i=1; i<=n; i++)//每一片的连通区域都要判断 

		if(!vis[i]) 

		{

			vis[i]=1;

			if(spfa(i)==false)  {cout<<"No";return 0;}//有约束自相矛盾 

		}

	cout<<"Yes";//所以的点都符合条件 

	return 0;

}

bool spfa(int s)

{

 	queue<int>q;

 	//memset(book,0,sizeof(book));

	memset(dis,0x7f,sizeof(dis));

	dis[s]=0;

	q.push(s);book[s]=1;	

	//如果没松弛就松弛，如果松弛过就判断是否符合约束条件，如果自相矛盾就是不可能的数据

	while(q.size()>0) 

	{

		int k=q.front();q.pop();//对头元素出队 

		book[k]=0;

		for(int i=head[k]; i!=-1; i=e[i].next)

		{

			int v=e[i].to;

			if(dis[v]==WQD && dis[k]<WQD)//如果v点没松弛过，就松弛更新dis[v]的值 				

			{	//题目的约束条件是= 			

				dis[v]=dis[k]+e[i].w;//更新 

				if(book[v]==0)	{q.push(v);	book[v]=1;}//入队标记

				vis[v]=1;//标记该点已经松弛过 

			}

			//据题意理解差分系统约束条件，不相等，自相矛盾 

			else if(dis[v]<WQD && dis[k]<WQD && dis[v]!=dis[k]+e[i].w)	return false;

		}		

    }

	return true;//别掉了这句 

}

// Faster C++ solution for the same problem (measured speedup: 10.573411000861581x):
// luogu-judger-enable-o2

// SeptEtavioxy

#include<cstdio>

#include<cctype>

#include<cstring>

#include<algorithm>

#include<cmath>

#define re register

#define ll long long

#define il inline

#define rep(i,s,t) for(re int i=(s);i<=(t);i++)

#define rev_rep(i,s,t) for(re int i=(s);i>=(t);i--)

#define each(i,u) for(int i=head[u];i;i=bow[i].nxt)

#define pt(ch) putchar(ch)

#define pti(x) printf("%d",x)

#define ptll(x) printf("%lld",x)

#define file(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)

using namespace std;

// c_ints

il int ci(){

	re char ch;int f=1;

	while(!isdigit(ch=getchar()))f= ch=='-'?-1:1 ;

	re int x= ch^'0';

	while(isdigit(ch=getchar()))x=(x*10)+(ch^'0');

	return f*x;

}

enum{N=100024};

int fa[N],dis[N];

void fd(int x){

	if( fa[x]==x ) return ;

	fd(fa[x]);

	dis[x]+= dis[fa[x]];

	fa[x]= fa[fa[x]];

}

int main(){

	int n=ci(), m=ci();

	rep(i,1,n) fa[i]= i;

	while( m-- ){

		int l=ci(), r=ci(), d=ci();

		fd(l), fd(r);

		if( fa[l]==fa[r] ){

			if( dis[r]-dis[l]!=d ){

				puts("No");

				return 0;

			}

		}

		else{

			int fl=fa[l], fr=fa[r];

			int x= dis[l]+d-dis[r];

//			printf("%d %d\n",fl,fr);

//			if( x<0 ){

//				puts("No");

//				return 0;

//			}

			dis[fr]= x;

			fa[fr]= fl;

		}

//		printf("fa :");rep(i,1,n) printf("%d ",fa[i]); pt('\n');

//		printf("dis:");rep(i,1,n) printf("%d ",dis[i]); pt('\n');

	}

	puts("Yes");

	return 0;

}