bzoj 2733: [HNOI2012]永无乡

传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=2733
思路:数据结构的启发式合并,把小的塞到大的里就好啦,,,复杂度Nlog^2N,不科学?很科学,,考虑每个点至多合并logn次,,所以,,,不多说,,用cin读入字符狂RE不止,,,我也是醉了,,,真的知道读入优化的必要了把,,,
代码:

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#define N 100000
using namespace std;
struct node { int data,sz,son[2],fa;};
node tr[N+5];
int n,m,f[N+5],w[N+5],que[N+5];
inline int find(int x){ return f[x] == x ? x : f[x] = find(f[x]);}
inline void update(int x){ tr[x].sz = tr[tr[x].son[0]].sz + tr[tr[x].son[1]].sz + 1;}
inline int getnum(){
    char c; int num;
    while (!isdigit(c = getchar()));
    num = c - '0';
    while (isdigit(c = getchar())) num = 10 * num + c - '0';
    return num;
}


void init(){
    n = getnum(); m = getnum();
    for (int i = 1;i <= n; ++i) w[i] = getnum();
    for (int i = 1;i <= n; ++i) f[i] = i;
    for (int i = 1;i <= n; ++i) tr[i].sz = 1, tr[i].data = w[i];
}

inline void rorate(int x){
     int y = tr[x].fa,z = tr[y].fa;
     bool p = (x == tr[y].son[1]),q = p^1;
     if (z) tr[z].son[(y==tr[z].son[1])] = x;
     tr[x].fa = z; tr[y].fa = x; tr[tr[x].son[q]].fa = y;
     tr[y].son[p] = tr[x].son[q]; tr[x].son[q] = y;
     update(y); update(x);
}

inline void splay(int x,int p){
    while (tr[x].fa != p){
        int y = tr[x].fa,z = tr[y].fa;
        if (z) 
          if ((y == tr[z].son[0])^(x == tr[y].son[0])) rorate(x);
          else rorate(y);
        rorate(x);
    }
    update(x);
}

inline void insert(int x,int root){
    int t = root,last = 0;
    while (t){
        last = t;
        if (tr[x].data > tr[t].data) t = tr[t].son[1];
        else t = tr[t].son[0];   
    }
    tr[x].fa = last; tr[last].son[(tr[x].data > tr[last].data)] = x;
    update(x); update(last);
    splay(x,0);
}

inline void merge(int u,int v){
    splay(u,0); splay(v,0);
    if (tr[u].sz > tr[v].sz) swap(u,v);
    f[u] = v;
    int head = 0,tail = 1,last = v;
    que[1] = u;
    while (head < tail){
        int x = que[++head];
        if (tr[x].son[0]) que[++tail] = tr[x].son[0];
        if (tr[x].son[1]) que[++tail] = tr[x].son[1];
    }
    for (int i = 1;i <= tail; ++i) insert(que[i],last),last = que[i];
}

inline int query(int t,int k){
    if (!t) return -1;
    int s = tr[tr[t].son[0]].sz + 1;
    if (s == k) return t;
    if (s > k) return query(tr[t].son[0],k);
    else query(tr[t].son[1],k - s);
}

void DO_IT(){
    int a,b,u1,v1;
    for (int i = 1;i <= m; ++i){
         a = getnum(),b = getnum();
         u1 = find(a),v1 = find(b);
        if (u1 != v1) merge(u1,v1);
    }
    int q = getnum();
    char ch;
    while (q--){
        ch = getchar();
        while (ch < 'A'||ch > 'Z') ch = getchar();
         a = getnum(),b = getnum();
        if (ch == 'B') {
            u1 = find(a),v1 = find(b);
            if (u1 != v1) merge(u1,v1);
        }
        else splay(a,0),printf("%d\n",query(a,b));
    }
}

int main(){
    init();
    DO_IT();
    return 0;
}

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