题目大意:给你n个点,每个点有权值k,现有两种操作:将两个点所在联通块合并,查询某个点所在联通块权值第k小是哪个数
Splay启发式合并解法在这里
这题输出的是点的编号不是点的值….
对于每个点,开个权值线段树,合并操作用并查集查找,线段树合并
查询操作是权值线段树基本操作,找第
k
小的数时,如果左儿子的数超过k个就在左儿子里,否则去右儿子找第
代码如下:
#include<ctype.h>
#include<cstdio>
#define N 100050
using namespace std;
const int INF=10000000;
inline int read(){
int x=0,f=1;char c;
do c=getchar(),f=c=='-'?-1:f; while(!isdigit(c));
do x=(x<<3)+(x<<1)+c-'0',c=getchar(); while(isdigit(c));
return x*f;
}
int n,m,x,y,fx,fy,t;
int f[N],pre[N];
char s[25];
struct Node{
int l,r,siz;
Node *ls,*rs;
Node();
inline void maintain(){
siz=ls->siz+rs->siz;
}
}*root[N],*null;
Node::Node(){
ls=rs=null;
}
int Find(int x){
return f[x]==x?f[x]:f[x]=Find(f[x]);
}
inline void init(){
null=new Node;
null->l=null->r=-1;
null->siz=0;
null->ls=null->rs=null;
}
void Merge(Node *&x,Node *&y){///线段树合并
if(y==null) return;
if(x==null){
x=y;
return;
}
x->siz+=y->siz;
Merge(x->ls,y->ls);Merge(x->rs,y->rs);
}
void Add(int x,int l,int r,Node *&k){
if(k==null){
k=new Node;
k->l=l;k->r=r;
}
if(k->l==k->r){k->siz++;return;}
int mid=k->l+k->r>>1;
if(x<=mid) Add(x,l,mid,k->ls);
else Add(x,mid+1,r,k->rs);
k->maintain();
}
int Rank(int x,Node *k){
if(k->siz<x) return -1;
if(k->l==k->r) return k->l;
if(x<=k->ls->siz) return Rank(x,k->ls);
else return Rank(x-k->ls->siz,k->rs);
}
void print(Node *x){
if(x==null) return;
printf("%d %d %d\n",x->l,x->r,x->siz);
print(x->ls);print(x->rs);
}
int main(){
init();
n=read();m=read();
for(int i=1;i<=n;i++){
f[i]=i;
x=read();
pre[x]=i;
Add(x,1,INF,root[i]=null);
}
for(int i=1;i<=m;i++){
x=read();y=read();
fx=Find(x),fy=Find(y);
if(fx==fy) continue;
Merge(root[fx],root[fy]);
f[fy]=fx;
}
m=read();
for(int i=1;i<=m;i++){
scanf("%s",s+1);
if(s[1]=='B'){
x=read();y=read();
fx=Find(x),fy=Find(y);
if(fx==fy) continue;
Merge(root[fx],root[fy]);
f[fy]=fx;
}
else{
x=read();y=read();
fx=Find(x);
t=Rank(y,root[fx]);
printf("%d\n",!~t?-1:pre[t]);
}
}
return 0;
}