BZOJ 3224: Tyvj 1728 普通平衡树

3224: Tyvj 1728 普通平衡树

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 9629  Solved: 4091
[Submit][Status][Discuss]

Description

您需要写一种数据结构（可参考题目标题），来维护一些数，其中需要提供以下操作：
1. 插入x数
2. 删除x数(若有多个相同的数，因只删除一个)
3. 查询x数的排名(若有多个相同的数，因输出最小的排名)
4. 查询排名为x的数
5. 求x的前驱(前驱定义为小于x，且最大的数)
6. 求x的后继(后继定义为大于x，且最小的数)

Input

第一行为n，表示操作的个数,下面n行每行有两个数opt和x，opt表示操作的序号(1<=opt<=6)

Output

对于操作3,4,5,6每行输出一个数，表示对应答案

Sample Input

10
1 106465
4 1
1 317721
1 460929
1 644985
1 84185
1 89851
6 81968
1 492737
5 493598

Sample Output

106465
84185
492737

HINT

1.n的数据范围：n<=100000

2.每个数的数据范围：[-1e7,1e7]

数据如下http://pan.baidu.com/s/1jHMJwO2

Source

平衡树

[Submit][Status][Discuss]

FHQ Treap

 #include <bits/stdc++.h>

 const int N = ;

 int ls[N], rs[N], vl[N], tg[N], sz[N];

 inline int node(int v)
 {
     static int t = ;
     sz[t] = ;
     vl[t] = v;
     tg[t] = rand();
     return t++;
 }

 int merge(int a, int b)
 {
     if (!a || !b)return a + b;
     if (tg[a] > tg[b])
     {
         rs[a] = merge(rs[a], b);
         sz[a] =  + sz[ls[a]] + sz[rs[a]];
         return a;
     }
     else
     {
         ls[b] = merge(a, ls[b]);
         sz[b] =  + sz[ls[b]] + sz[rs[b]];
         return b;
     }
 }

 void split(int t, int k, int &a, int &b)
 {
     if (!t)a = b = ;
     else
     {
         if (vl[t] <= k)
             a = t, split(rs[t], k, rs[t], b);
         else
             b = t, split(ls[t], k, a, ls[t]);
         sz[t] =  + sz[ls[t]] + sz[rs[t]];
     }
 }

 int kth(int t, int k)
 {
     if (k <= sz[ls[t]])
         return kth(ls[t], k);
     else if (k == sz[ls[t]] + )
         return t;
     else
         return kth(rs[t], k - sz[ls[t]] - );
 }

 signed main(void)
 {
     srand();

     int n, r = ; scanf("%d", &n);

     for (int a, b, x, y, z; n--; )
     {
         scanf("%d%d", &a, &b);

         if (a == )
         {
             split(r, b, x, y);
             r = merge(x, node(b));
             r = merge(r, y);
         }
         else if (a == )
         {
             split(r, b, x, z);
             split(x, b - , x, y);
             y = merge(ls[y], rs[y]);
             r = merge(x, y);
             r = merge(r, z);
         }
         else if (a == )
         {
             split(r, b - , x, y);
             printf("%d\n", sz[x] + );
             r = merge(x, y);
         }
         else if (a == )
             printf("%d\n", vl[kth(r, b)]);
         else if (a == )
         {
             split(r, b - , x, y);
             printf("%d\n", vl[kth(x, sz[x])]);
             r = merge(x, y);
         }
         else
         {
             split(r, b, x, y);
             printf("%d\n", vl[kth(y, )]);
             r = merge(x, y);
         }
     }
 }  

 #include <bits/stdc++.h>
 const int N = ;
 int ls[N], rs[N], vl[N], tg[N], sz[N], tot = ;
 int node(int v) {
     return vl[tot] = v, sz[tot] = , tg[tot] = rand(), tot++;
 }
 int merge(int a, int b) {
     if (!a || !b)return a + b;
     if (tg[a] > tg[b]) {
         rs[a] = merge(rs[a], b);
         sz[a] =  + sz[ls[a]] + sz[rs[a]];
         return a;
     }
     else {
         ls[b] = merge(a, ls[b]);
         sz[b] =  + sz[ls[b]] + sz[rs[b]];
         return b;
     }
 }
 int split(int t, int k, int &a, int &b) {
     if (!t)return a = b = , ;
     if (vl[t] <= k)
         a = t, split(rs[t], k, rs[t], b);
     else
         b = t, split(ls[t], k, a, ls[t]);
     return sz[t] =  + sz[ls[t]] + sz[rs[t]];
 }
 int kth(int t, int k) {
     return k <= sz[ls[t]] ? kth(ls[t], k) : ((k -= sz[ls[t]] + ) ? kth(rs[t], k) : vl[t]);
 }
 signed main(void) {
     int n, r = , a, b, x, y, z;
     for (scanf("%d", &n); n--; ) {
         scanf("%d%d", &a, &b);
         if (a == )
             split(r, b, x, y), r = merge(merge(x, node(b)), y);
         else if (a == )
             split(r, b, x, z), split(x, b - , x, y), r = merge(merge(x, merge(ls[y], rs[y])), z);
         else if (a == )
             split(r, b - , x, y), printf("%d\n", sz[x] + ), r = merge(x, y);
         else if (a == )
             printf("%d\n", kth(r, b));
         else if (a == )
             split(r, b - , x, y), printf("%d\n", kth(x, sz[x])), r = merge(x, y);
         else
             split(r, b, x, y), printf("%d\n", kth(y, )), r = merge(x, y);
     }
 }

@Author: YouSiki