/* 题意:u,v,w队长,队员,队长留下两个队员可以回家,两个队员留下,队长回家 2-sat问题,把两个队员看成一个整体就变成一个简单2-sat问题了 */ #include<stdio.h> #include<string.h> #include<algorithm> #include<stdlib.h> using namespace std; #define N 6100 #define NN 5100 struct node { int u,v,w,next; }bian[NN*8]; int head[N],yong,low[N],dfn[N],belong[N],ans,top,index,stac[N],vis[N]; void init() { memset(head,-1,sizeof(head)); yong=index=ans=top=0; memset(vis,0,sizeof(vis)); memset(dfn,0,sizeof(dfn)); } void addedge(int u,int v) { bian[yong].v=v; bian[yong].next=head[u]; head[u]=yong++; } void tarjan(int u) { low[u]=dfn[u]=++index; stac[++top]=u; vis[u]=1; int i; for(i=head[u]; i!=-1; i=bian[i].next) { int v=bian[i].v; if(!dfn[v]) { tarjan(v); low[u]=min(low[u],low[v]); } else if(vis[v]) low[u]=min(low[u],dfn[v]); } if(low[u]==dfn[u]) { ans++; int t; do { t=stac[top--]; belong[t]=ans; vis[t]=0; } while(t!=u); } } int slove(int n) { int i; for(i=0; i<=n; i++) if(!dfn[i]) tarjan(i); // printf("%d ",ans); for(i=0; i<=n; i+=2) if(belong[i]==belong[i+1]) return 0; return 1; } int indx[N]; int main() { int n,m,i,u,len,v,w; while(scanf("%d%d",&n,&m)!=EOF) { init(); len=-1; for(i=0;i<n;i++) { scanf("%d%d%d",&u,&v,&w); indx[u]=++len; indx[v]=++len; indx[w]=len; } for(i=0;i<m;i++) { scanf("%d%d",&u,&v); addedge(indx[u],indx[v]^1); addedge(indx[v],indx[u]^1); } if(!slove(len)) printf("no "); else printf("yes "); } return 0;}