[HNOI2012]永无乡 SBT+启发式合并

题目描述
题目

我也不知道启发式合并到底是什么东西
好吧，启发式合并就是把小的集合往大的集合合并，学并查集的时候不是有把小树根的父亲设为大树根的操作吗，这大概就是启发式合并的一个典例

对于这道题，我的做法是把小SBT的所有结点从小到大全部加入到大SBT中（本来打了个Treap，懒得旋转了，就改成了SBT，还好没有极限数据卡我），大概把并查集的操作省略了

代码

#include<iostream>#include<cstdio>#include<cstdlib>#include<algorithm>#include<cstring>using namespace std;const int N=100010;int n, q, w[N], m;struct node{    node* ch[2];    int v, id, s, r;    node(int id, int v):id(id), v(v) {ch[0]=ch[1]=NULL; s=1; r=rand();}    void maintain()    {        s=1;        if(ch[0] != NULL) s+=ch[0]->s;        if(ch[1] != NULL) s+=ch[1]->s;    }};node* rt[N];//rt[i]记的是编号为i的结点所在SBT的根void ins(node* &o, int x, int v){    if(o == NULL) {o=new node(x, v); return ;}    int d=(v > o->v); ins(o->ch[d], x, v);    o->maintain();}void dfs(node* &o, node* &k){    if(k == NULL) return ;    dfs(o, k->ch[0]);    ins(o, k->id, k->v);    dfs(o, k->ch[1]);    delete k;}void merge(int a, int b){    if(rt[a] == NULL || rt[b] == NULL) return;    if(rt[a]->s < rt[b]->s) swap(a, b);    dfs(rt[a], rt[b]);    rt[b]=rt[a];}int find(node* o, int k){    if(o == NULL || k <= 0 || k > o->s) return -1;    int ss=(o->ch[0] == NULL ? 0 : o->ch[0]->s);    if(k == ss+1) return o->id;    if(k <= ss) return find(o->ch[0], k);    return find(o->ch[1], k-ss-1);}int read(){    int out=0;char c=getchar();while(c < '0' || c > '9') c=getchar();    while(c >= '0' && c <= '9') out=(out<<1)+(out<<3)+c-'0',c=getchar();return out;}void solve(){    n=read(), m=read();int u, v;    for(int i=1; i <= n; i++) w[i]=read(), ins(rt[i], i, w[i]);    for(int i=1; i <= m; i++) {u=read(), v=read(); merge(u, v);}    q=read();    for(int i=1; i <= q; i++)     {        char c=getchar();        while(c != 'B' && c != 'Q') c=getchar();        if(c == 'B')         {            u=read(), v=read();            if(u > n || u < 1 || v > n || v <1) continue;            if(rt[u] != rt[v]) merge(u, v);        }        else         {            u=read(), v=read();            if(u > n || u < 1) {printf("-1\n"); continue;}            int ans=find(rt[u], v); printf("%d\n",ans);        }    }}int main(){    solve();    return 0;}