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;
}
写个下动态开点的权值线段树，为啥跑出来比splay还要慢啊！这么玄学的吗。。
#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;
int tree[N<<5], root[N], lc[N<<5], rc[N<<5], pos[N], parent[N];
int find(int p){
while(p != parent[p]) parent[p] = parent[parent[p]], p = parent[p];
return p;
}
int build(){
++tot;
tree[tot] = lc[tot] = rc[tot] = 0;
return tot;
}
void push_up(int x){
tree[x] = tree[lc[x]] + tree[rc[x]];
}
void modify(int rt, int l, int r, int k){
if(l == r){
tree[rt] ++;
return;
}
int mid = (l + r) >> 1;
if(k <= mid){
if(!lc[rt]) lc[rt] = build();
modify(lc[rt], l, mid, k);
}
else{
if(!rc[rt]) rc[rt] = build();
modify(rc[rt], mid + 1, r, k);
}
push_up(rt);
}
int merge(int x, int y, int l, int r){
if(!x) return y;
if(!y) return x;
if(l == r){
tree[x] += tree[y];
return x;
}
int mid = (l + r) >> 1;
lc[x] = merge(lc[x], lc[y], l, mid);
rc[x] = merge(rc[x], rc[y], mid + 1, r);
push_up(x);
return x;
}
int query(int rt, int l, int r, int k){
if(l == r) return l;
int mid = (l + r) >> 1;
if(k <= tree[lc[rt]]) return query(lc[rt], l, mid, k);
return query(rc[rt], mid + 1, r, k - tree[lc[rt]]);
}
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();
pos[v] = i, root[i] = build(), modify(root[i], 1, n, v);
}
for(int i = 0; i < m; i ++){
int x = read(), y = read();
int fx = find(x), fy = find(y);
if(fx == fy) continue;
parent[fy] = fx;
root[fx] = merge(root[fx], root[fy], 1, n);
}
int q = read();
while(q --){
char opt[10]; scanf("%s", opt);
if(opt[0] == 'B'){
int x = read(), y = read();
int fx = find(x), fy = find(y);
if(fx == fy) continue;
parent[fy] = fx;
root[fx] = merge(root[fx], root[fy], 1, n);
}
else if(opt[0] == 'Q'){
int x = read(), k = read();
if(tree[root[find(x)]] < k) printf("-1\n");
else printf("%d\n", pos[query(root[find(x)], 1, n, k)]);
}
}
return 0;
}
来源：https://www.cnblogs.com/onionQAQ/p/10771171.html