splay启发式合并

启发式合并其实就是把集合数量小的合并到集合数量大的里去。
怎么合并呢，直接一个一个插入就行了。。

用并查集维护连通性，find(i)可以找到所在splay的编号

这题好像还可以合并线段树来写，下次再补上。。

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 100005;
int n, m, tot, fa[N<<5], ch[N<<5][2], size[N<<5], val[N<<5], root[N], parent[N], pos[N];

int find(int p){
    while(p != parent[p]) parent[p] = parent[parent[p]], p = parent[p];
    return p;
}

bool isConnect(int p, int q){
    return find(p) == find(q);
}

int init(int v, int f){
    ++tot;
    size[tot] = 1, val[tot] = v, fa[tot] = f;
    ch[tot][0] = ch[tot][1] = 0;
    return tot;
}

void push_up(int x){
    size[x] = size[ch[x][0]] + size[ch[x][1]] + 1;
}

void rotate(int x){
    int y = fa[x], z = fa[y], p = (ch[y][1] == x) ^ 1;
    ch[y][p^1] = ch[x][p], fa[ch[x][p]] = y;
    ch[z][ch[z][1] == y] = x, fa[x] = z;
    ch[x][p] = y, fa[y] = x;
    push_up(y), push_up(x);
}

void splay(int rt, int x, int goal){
    if(x == goal) return;
    while(fa[x] != goal){
        int y = fa[x], z = fa[y];
        if(z != goal){
            (ch[y][0] == x) ^ (ch[z][0] == y) ? rotate(x) : rotate(y);
        }
        rotate(x);
    }
    push_up(x);
    if(goal == 0) root[rt] = x;
}

int select(int rt, int k){
    if(!root[rt]) return 0;
    int cur = root[rt], p = size[ch[root[rt]][0]];
    while(1){
        if(p + 1 < k){
            k -= p + 1;
            cur = ch[cur][1];
        }
        else{
            if(p + 1 == k) return val[cur];
            cur = ch[cur][0];
        }
        p = size[ch[cur][0]];
    }
}

void insert(int rt, int x){
    if(!root[rt]){
        root[rt] = init(x, 0);
        return;
    }
    int cur = root[rt];
    while(ch[cur][x > val[cur]]){
        if(val[cur] == x) break;
        cur = ch[cur][x > val[cur]];
    }
    if(val[cur] == x){
        splay(rt, cur, 0);
        return;
    }
    ch[cur][x > val[cur]] = init(x, cur);
    splay(rt, ch[cur][x > val[cur]], 0);
}

void dfs(int x, int y){
    if(!x) return;
    dfs(ch[x][0], y);
    dfs(ch[x][1], y);
    insert(y, val[x]);
}

void merge(int x, int y){
    int fx = find(x), fy = find(y);
    if(fx == fy) return;
    if(size[root[fx]] > size[root[fy]]) swap(fx, fy);
    parent[fx] = fy;
    dfs(root[fx], fy);
}

int main(){

    n = read(), m = read();
    for(int i = 0; i <= n; i ++) parent[i] = i;
    for(int i = 1; i <= n; i ++){
        int v = read();
        root[i] = init(v, 0);
        pos[v] = i;
    }
    for(int i = 0; i < m; i ++){
        int x = read(), y = read();
        if(isConnect(x, y)) continue;
        merge(x, y);
    }
    int q = read();
    while(q --){
        char opt[5]; scanf("%s", opt);
        int x = read(), y = read();
        if(opt[0] == 'B'){
            if(isConnect(x, y)) continue;
            merge(x, y);
        }
        else if(opt[0] == 'Q'){
            if(size[root[find(x)]] < y) printf("-1\n");
            else printf("%d\n", pos[select(find(x), y)]);
        }
    }
    return 0;
}

标签：ch,return,cur,val,2733,int,永无,fx,BZOJ
来源： https://www.cnblogs.com/onionQAQ/p/10771171.html