Acwing1063 永无乡 题解

题面

传送门

题解

并查集+Splay+启发式合并
启发式合并：
每次合并两个Splay时，将节点数小的合并至节点数大的。
神奇的时间复杂度：完成所有的合并总共\(O(N\log{N})\)，然而不会证。此题合并平衡树，则为\(O(N\log^2{N})\)。
其他没什么了。

Code

#include<cstdio>
#include<algorithm>
using namespace std;
const int N = 1e5 + 5;
struct Splay {
    int s[2], p, val, size;
    void init(int _v, int _p) {
        val = _v; p = _p; size = 1;
    }
} a[N];
int n, m, r[N], root[N];
int fa[N], sz[N];
int get(int x) {
    if (x == fa[x]) return x;
    return fa[x] = get(fa[x]);
}
void pushup(int x) {
    a[x].size = a[a[x].s[0]].size + a[a[x].s[1]].size + 1;
}
void rotate(int x) {
    int y = a[x].p, z = a[y].p, k = a[y].s[1] == x;
    a[z].s[a[z].s[1] == y] = x; a[x].p = z;
    a[y].s[k] = a[x].s[k ^ 1]; a[a[x].s[k ^ 1]].p = y;
    a[x].s[k ^ 1] = y; a[y].p = x;
    pushup(y); pushup(x);
}
void Splay(int x, int k, int rt) {
    while (a[x].p != k) {
        int y = a[x].p, z = a[y].p;
        if (z != k) {
            if (a[y].s[1] == y ^ a[y].s[1] == x) rotate(x);
            else rotate(y);
        }
        rotate(x);
    }
    if (!k) root[rt] = x;
}
void Insert(int x, int rt) {
    int p = root[rt], q = 0;
    while (p) {
        q = p; p = a[p].s[a[x].val > a[p].val];
    }
    a[x].init(a[x].val, q);
    a[q].s[a[x].val > a[q].val] = x;
    Splay(x, 0, rt);
}
void Transfer(int p, int rt) {
    if (!p) return ;
    Transfer(a[p].s[0], rt); Transfer(a[p].s[1], rt);
    Insert(p, rt);
}
void Merge(int x, int y) {
    if (sz[x] > sz[y]) swap(x, y);
    fa[x] = y; sz[y] += sz[x];
    Transfer(root[x], y);
}
int Get_K(int x, int k) {
    int p = root[x];
    while (p) {
        if (a[a[p].s[0]].size >= k) p = a[p].s[0];
        else if (a[a[p].s[0]].size + 1 == k) return p;
        else {
            k -= a[a[p].s[0]].size + 1; p = a[p].s[1];
        }
    }
    return -1;
}
int main() {
    char opt[5];
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &r[i]);
        a[i].init(r[i], 0);
        fa[i] = i; sz[i] = 1;
        root[i] = i;
    }
    while (m--) {
        int x, y;
        scanf("%d%d", &x, &y);
        Merge(x, y);
    }
    int q;
    scanf("%d", &q);
    while (q--) {
        int x, y;
        scanf("%s%d%d", opt, &x, &y);
        if (opt[0] == 'B') Merge(get(x), get(y));
        else printf("%d\n", Get_K(get(x), y));
    }
    return 0;
}

标签：rt,return,val,int,题解,永无,void,Acwing1063,size
来源： https://www.cnblogs.com/Fisher-Y/p/16036859.html