n个节点，m个询问，root为根节点，fa[i][j]表示i的第2^j个祖先

#include
     
      using namespace std;const int P=30;//vector son[500001];//若清楚每条边哪个是父节点那个是子节点，可以用vector，son[i]表示以i为父节点的子节点；int n,m,deep[500001],fa[500001][30]={0},root,jump[500005],num=0;bool judge[500005]={0};struct nob{	int sta,ed,jump;}a[1000005];//邻接表存储边void add(int sta,int ed){	num++;	a[num].sta=sta;	a[num].ed=ed;	a[num].jump=jump[sta];	jump[sta]=num;}void dfs(int pos){	judge[pos]=1;	for (int i=jump[pos]; i>0; i=a[i].jump){		if (judge[a[i].ed]) continue;		deep[a[i].ed]=deep[pos]+1;		fa[a[i].ed][0]=pos;		dfs(a[i].ed);	}}void beizeng(){	for (int i=1; i
      
       deep[x])	swap(x,y);	int s=deep[x]-deep[y];	for (int i=0; i
       
        =0; i--){		if (fa[x][i]!=fa[y][i]){			x=fa[x][i];			y=fa[y][i];		}	}	return fa[x][0];}int main(){	scanf("%d%d%d",&n,&m,&root);	//若没有说明根节点的话，那么fa[i][0]==0的为根节点 	for (int i=1,x,y; i