[Bzoj3224][Tyvj1728] 普通平衡树(splay/无旋Treap)

题目链接：https://www.lydsy.com/JudgeOnline/problem.php?id=3224

平衡树入门题，学习学习。

splay(学习yyb巨佬)

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int maxn = ;
 const int INF = 1e9;
 inline void read(int &n) {
     register int x = , t = ;
     register char ch = getchar();
     while (ch != '-' && (ch<'' || ch>''))ch = getchar();
     if (ch == '-') { t = -; ch = getchar(); }
     while (ch >= ''&&ch <= '') { x = x *  + ch - ; ch = getchar(); }
     n = x * t;
 }
 int root, len;
 struct node {
     int fa;//节点的父节点
     int siz;//节点的子树数量
     int ch[];//节点的左(0)子节点和右(1)子节点
     int val;//节点的值
     int num;//节点的值的数量
 }a[maxn];
 inline void pushup(int x) {//更新当前节点的子树//当前子树的大小是左子树大小加上右子树大小当前当前节点个数
     a[x].siz = a[a[x].ch[]].siz + a[a[x].ch[]].siz + a[x].num;
 }
 inline void rotate(int x) {//旋转操作//将x旋转到x的父亲的位置
     int y = a[x].fa;//y是x的父节点
     int z = a[y].fa;//z是x的祖父节点
     int k = (a[y].ch[] == x);//求x是y的左(0)还是右(1)节点
     a[z].ch[a[z].ch[] == y] = x;//求y是z的左(0)还是右(1)节点，同时将x放在该位置
     a[x].fa = z;//修改x的父节点
     a[y].ch[k] = a[x].ch[k ^ ];//把x的儿子给y
     a[a[x].ch[k ^ ]].fa = y;//更新父节点为y
     a[x].ch[k ^ ] = y;//y变为x的子节点
     a[y].fa = x;//y的父亲更新为x
     pushup(y), pushup(x);//更新y和x的子节点数量
 }
 inline void splay(int x, int goal) {//将x旋转为goal的子节点
     while (a[x].fa != goal) {
         int y = a[x].fa;
         int z = a[y].fa;
         if (z != goal)
             (a[y].ch[] == x) ^ (a[z].ch[] == y) ? rotate(x) : rotate(y);
         //如果x和y同为左儿子或者右儿子先旋转y
         //如果x和y不同为左儿子或者右儿子先旋转x
         //如果不双旋的话，旋转完成之后树的结构不会变化
         rotate(x);
     }
     if (goal == )
         root = x;
 }
 inline void insert(int x) {
     int now = root, fa = ;//当前节点now和now的父节点fa
     while (now&&a[now].val != x) {//当now存在并且没有移动到当前的值
         fa = now;//父节点变为now
         now = a[now].ch[x > a[now].val];//x大于当前位置则now向右，否则now向左
     }
     if (now)//这个值出现过
         a[now].num++;//值增加
     else {//这个值第一次出现，要新建一个节点来存放
         now = ++len;
         if (fa)//父节点不是根节点(0)
             a[fa].ch[x > a[fa].val] = now;
         //初始化节点
         a[now].ch[] = a[now].ch[] = ;
         a[now].siz = ;
         a[now].num = ;
         a[now].fa = fa;
         a[now].val = x;
     }
     splay(now, );
 }
 inline void  Find(int x) {//查找x的位置，并将其旋转到根节点
     int now = root;
     if (!now)return;//数为空
     while (a[now].ch[x > a[now].val] && x != a[now].val)////当存在儿子并且当前位置的值不等于x
         now = a[now].ch[x > a[now].val];
     splay(now, );//把当前位置旋转到根节点
 }
 inline int Next(int x, int flag) {//查找x的前驱(0)或者后继(1)
     Find(x);
     int now = root;
     if (a[now].val > x&&flag)return now;//如果当前节点的值大于x并且要查找的是后继
     if (a[now].val < x && !flag)return now;//如果当前节点的值小于x并且要查找的是前驱
     now = a[now].ch[flag];//查找后继的话在右儿子上找，前驱在左儿子上找
     while (a[now].ch[flag ^ ])now = a[now].ch[flag ^ ];
     return now;
 }
 inline void Delete(int x) {//删除一个x
     int last = Next(x, );//查找x的前驱
     int next = Next(x, );//查找x的后继
     splay(last, ), splay(next, last);
     //将前驱旋转到根节点，后继旋转到根节点下面
     //很明显，此时后继是前驱的右儿子，x是后继的左儿子，并且x是叶子节点
     int now = a[next].ch[];//后继的左儿子，也就是x
     if (a[now].num > ) {//超过一个就删除一个
         a[now].num--;
         splay(now, );
     }
     else
         a[next].ch[] = ;//直接删除

 }
 inline int kth(int k) {//查找第k小的数
     int now = root;
     if (a[now].siz < k)
         return ;
     while () {
         int y = a[now].ch[];//y是左儿子
         if (k > a[y].siz + a[now].num) {//如果排名比左儿子的大小和当前节点的数量要大
             k -= a[y].siz + a[now].num;//排名减小
             now = a[now].ch[];//定在右儿子上找
         }
         else {
             if (k <= a[y].siz)
                 now = y;
             else
                 return a[now].val;
         }
     }
 }
 int main() {
     int n;
     read(n);
     root = , len = ;
     insert();
     insert(-);
     while (n--) {
         int q, x;
         read(q), read(x);
         if (q == )
             insert(x);
         else if (q == )
             Delete(x);
         else if (q == ) {
             Find(x);
             printf("%d\n", a[a[root].ch[]].siz);
         }
         else if (q == )
             printf("%d\n", kth(x + ));
         else if (q == )
             printf("%d\n", a[Next(x, )].val);
         else if (q == )
             printf("%d\n", a[Next(x, )].val);
     }
 }

无旋Treap(学习fzszkl巨佬)

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int maxn = 1e5 + ;
 int Siz[maxn], ls[maxn], rs[maxn], val[maxn], pos[maxn], root;
 int cnt;
 int send = ;
 inline int Rand() {
     send ^= send << ;
     send ^= send >> ;
     send ^= send << ;
     return send;
 }
 inline int New(int x) {
     cnt++;
     pos[cnt] = Rand();
     Siz[cnt] = ;
     val[cnt] = x;
     return cnt;
 }
 inline void up(int x) {
     Siz[x] = Siz[ls[x]] + Siz[rs[x]] + ;
 }
 void split_size(int x, int siz, int &A, int &B) {
     if (x == )return (void)(A = B = );
     if (siz <= Siz[ls[x]])
         B = x, split_size(ls[x], siz, A, ls[x]);
     else
         A = x, split_size(rs[x], siz - Siz[ls[x]] - , rs[x], B);
     up(x);
 }
 void split_val(int x, int v, int &A, int &B) {
     if (x == )return (void)(A = B = );
     if (v < val[x])
         B = x, split_val(ls[x], v, A, ls[x]);
     else
         A = x, split_val(rs[x], v, rs[x], B);
     up(x);
 }
 int Merge(int A, int B) {
     if (A ==  || B == )return A | B;
     int ans;
     if (pos[A] > pos[B])ans = A, rs[A] = Merge(rs[A], B);
     else ans = B, ls[B] = Merge(A, ls[B]);
     up(ans);
     return ans;
 }
 void dfs(int x) {
     if (x == )
         return;
     printf("%d ", x);
     dfs(ls[x]);
     dfs(rs[x]);
 }
 void insert(int x) {
     int A, B;
     split_val(root, x - , A, B);
     root = Merge(Merge(A, New(x)), B);
 }
 void Delete(int x) {
     int A, B, C, D;
     split_val(root, x - , A, B);
     split_size(B, , C, D);//C就是删除的节点
     root = Merge(A, D);
 }
 int Rank(int x) {//查x的排名
     int A, B, ans;
     split_val(root, x - , A, B);
     ans = Siz[A] + ;
     root = Merge(A, B);
     return ans;
 }
 int getR(int x) {//查第x名是什么
     int A, B, C, D, ans;
     split_size(root, x - , A, B);
     split_size(B, , C, D);
     ans = val[C];
     Merge(Merge(A, C), D);
     return ans;
 }
 int last(int x) {
     int A, B, C, D, ans;
     split_val(root, x - , A, B);
     split_size(A, Siz[A] - , C, D);
     ans = val[D];
     Merge(Merge(C, D), B);
     return ans;
 }
 int Next(int x) {
     int A, B, C, D, Ans;
     split_val(root, x, A, B);
     split_size(B, , C, D);
     Ans = val[C];
     root = Merge(Merge(A, C), D);
     return Ans;
 }
 int main() {
     int n;
     scanf("%d", &n);
     while (n--) {
         int opt, x;
         scanf("%d%d", &opt, &x);
         if (opt == )
             insert(x);
         else if (opt == )
             Delete(x);
         else if (opt == )
             printf("%d\n", Rank(x));
         else if (opt == )
             printf("%d\n", getR(x));
         else if (opt == )
             printf("%d\n", last(x));
         else if (opt == )
             printf("%d\n", Next(x));
     }
     return ;
 }