bzoj 2733 永无乡 线段树

题目:

支持两种操作:

1. 合并两点所在的联通块
2. 查询某点所在联通块内权值第k小.

题解

平衡树启发式合并随便搞一搞就好了。

我写了一个线段树合并

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
inline void read(int &x){
    x=0;char ch;bool flag = false;
    while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;
    while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
const int maxn = 100010;
struct Node{
    Node *ch[2];
    int siz,id;
    void update(){
	siz = ch[0]->siz + ch[1]->siz;
    }
}*null,mem[maxn*30],*root[maxn],*it;
inline void init(){
    it = mem;null = it++;
    null->ch[0] = null->ch[1] = null;
    null->siz = 0;
}
inline Node* newNode(){
    Node *p = it++;p->ch[0] = p->ch[1] = null;
    p->siz = 0;return p;
}
inline void insert(Node* &p,int l,int r,int pos,int id){
    if(p == null) p = newNode();
    if(l == r){
	p->siz ++ ;
	p->id = id;
	return ;
    }
    int mid = l+r >> 1;
    if(pos <= mid) insert(p->ch[0],l,mid,pos,id);
    else insert(p->ch[1],mid+1,r,pos,id);
    p->update();return ;
}
inline Node* Union(Node *x,Node *y){
    if(x == null) return y;
    if(y == null) return x;
    x->ch[0] = Union(x->ch[0],y->ch[0]);
    x->ch[1] = Union(x->ch[1],y->ch[1]);
    x->update();return x;
}
int n;
inline int query(Node *p,int k){
    if(k < 1 || k > p->siz) return -1;
    int l = 1,r = n;
    while(1){
	if(l == r) return p->id;
	int mid = l+r >> 1;
	if(p->ch[0]->siz >= k){
	    p = p->ch[0];
	    r = mid;
	}else{
	    k -= p->ch[0]->siz;
	    p = p->ch[1];
	    l = mid+1;
	}
    }
}
int fa[maxn];
inline int find(int x){
    return fa[x] == x ? x : fa[x] = find(fa[x]);
}
int main(){
    init();int m;read(n);read(m);
    for(int i=1;i<=n;++i) root[i] = null,fa[i] = i;
    for(int i=1,x;i<=n;++i){
	read(x);
	insert(root[i],1,n,x,i);
    }
    for(int i=1,u,v;i<=m;++i){
	read(u);read(v);
	int x = find(u);
	int y = find(v);
	if(x == y) continue;
	fa[x] = y;
	root[y] = Union(root[x],root[y]);
    }
    int q;read(q);
    char ch;
    int x,k,u,v;
    while(q--){
	while(ch=getchar(),ch<'!');
	if(ch == 'Q'){
	    read(x);read(k);
	    int fx = find(x);
	    printf("%d\n",query(root[fx],k));
	}else if(ch == 'B'){
	    read(u);read(v);
	    int x = find(u);
	    int y = find(v);
	    if(x == y) continue;
	    fa[x] = y;
	    root[y] = Union(root[x],root[y]);
	}
    }
    return 0;
}