解析

关键性质是一个定理：若m>3*n-6，必然不存在合法的平面图
~~这谁知道啊~~
不过这题应该往也许图过于稠密时必然无解这方面想
所以我们只需要考虑m、n同阶的情况就行了
这个时候我们直接暴力判断跑2-SAT就行了

代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=35605;
const int M=30005;
const int mod=998244353;
const double eps=1e-6;
ll read(){ll x=0,f=1;char c=getchar();while(!isdigit(c)){if(c=='-') f=-1;c=getchar();}while(isdigit(c)){x=x*10+c-'0';c=getchar();}return x*f;
}
int n,m;
struct node{int to,nxt;
}p[2005000];
int fi[N],cnt;
void addline(int x,int y){p[++cnt]=(node){y,fi[x]};fi[x]=cnt;//printf("x=%d y=%d\n",x,y);
}
int u[M],v[M],pos[N];
bool judge(int i,int j){return (u[i]<u[j]&&v[i]>u[j]&&v[i]<v[j])||(v[i]>v[j]&&u[i]>u[j]&&u[i]<v[j]);
}
int dfn[N],low[N],zhan[N],col[N],tim,tot,top;
void tarjan(int x){dfn[x]=low[x]=++tim;zhan[++top]=x;for(int i=fi[x];~i;i=p[i].nxt){int to=p[i].to;if(!dfn[to]){tarjan(to);low[x]=min(low[x],low[to]);}else if(!col[to]) low[x]=min(low[x],dfn[to]);}if(low[x]==dfn[x]){col[x]=++tot;while(zhan[top]!=x){col[zhan[top--]]=tot;}top--;}return;
}
void init(){memset(fi,-1,sizeof(fi));cnt=-1;memset(dfn,0,sizeof(dfn));memset(low,0,sizeof(low));memset(col,0,sizeof(col));memset(zhan,0,sizeof(zhan));tim=top=0;tot=2*m;
}
int main(){int T=read();while(T--){n=read();m=read();init();for(int i=1;i<=m;i++){u[i]=read(),v[i]=read();}for(int i=1;i<=n;i++){int x=read();pos[x]=i;}if(m>3*n-6){printf("NO\n");continue;}for(int i=1;i<=m;i++){u[i]=pos[u[i]];v[i]=pos[v[i]];if(u[i]>v[i]) swap(u[i],v[i]);//printf("%d: (%d %d)\n",i,u[i],v[i]);}for(int i=1;i<=m;i++){//if(v[i]==u[i]+1) continue;for(int j=i+1;j<=m;j++){//if(v[j]==u[j]+1) continue;if(judge(i,j)){addline(i,j+m);addline(j,i+m);addline(j+m,i);addline(i+m,j);//printf("i=%d j=%d ok\n",i,j);}}}for(int i=1;i<=m*2;i++){if(!dfn[i]) tarjan(i);}int flag=1;for(int i=1;i<=m;i++){if(col[i]==col[i+m]){flag=0;break;}}if(flag) printf("YES\n");else printf("NO\n");}
}
/*
2
1 4
*/