SPOJ ORDERSET Order statistic set 简单平衡树 或 树状数组

题目：

http://www.spoj.com/problems/ORDERSET/en/

题意：

有下面四种操作：

• I x 往集合中插入x，若存在则不操作
• D x 从集合中删除x，若不存在则不操作
• K x 求集合中第x大的数，若x大于集合的大小输出invalid
• C x 统计集合中小于x的数的个数

思路：

简单的平衡树题目，用treap很容易实现，另外学了一波pb_ds试着用了一下
treap：

#include <bits/stdc++.h>using namespace std;const int N = 200000 + 10, INF = 0x3f3f3f3f;struct node{    int val, pri, sz, son[2];    void init(int _val, int _pri, int _sz)    {        val = _val, pri = _pri, sz = _sz;        son[0] = son[1] = 0;    }}tr[N];int treap_root, treap_tot;void treap_init(){    treap_root = treap_tot = 0;    tr[0].init(0, 0, 0);}void treap_update(int x){    tr[x].sz = tr[tr[x].son[0]].sz + tr[tr[x].son[1]].sz + 1;}void treap_rotate(int &x, int p){    int y = tr[x].son[!p];    tr[x].son[!p] = tr[y].son[p];    tr[y].son[p] = x;    treap_update(x); treap_update(y);    x = y;}void treap_insert(int &x, int val){    if(! x) tr[x = ++treap_tot].init(val, rand(), 1);    else    {        tr[x].sz++;        int p = val > tr[x].val;        treap_insert(tr[x].son[p], val);        if(tr[x].pri < tr[tr[x].son[p]].pri) treap_rotate(x, !p);    }}bool treap_find(int x, int val){    if(! x) return false;    if(tr[x].val == val) return true;    int p = val > tr[x].val;    return treap_find(tr[x].son[p], val);}void treap_del(int &x, int val){    if(tr[x].val == val)    {        if(tr[x].son[0] && tr[x].son[1])        {            int p  = tr[tr[x].son[0]].pri > tr[tr[x].son[1]].pri;            treap_rotate(x, p);            treap_del(x, val);        }        else x = tr[x].son[0] + tr[x].son[1];    }    else    {        tr[x].sz--;        int p = val > tr[x].val;        treap_del(tr[x].son[p], val);    }}int treap_kth(int x, int k){    if(k == tr[tr[x].son[0]].sz + 1) return tr[x].val;    else if(k > tr[tr[x].son[0]].sz + 1) return treap_kth(tr[x].son[1], k - tr[tr[x].son[0]].sz - 1);    else return treap_kth(tr[x].son[0], k);}int treap_rank(int x, int val){    if(! x) return 0;    if(val == tr[x].val) return tr[tr[x].son[0]].sz;    else if(val > tr[x].val) return tr[tr[x].son[0]].sz + 1 + treap_rank(tr[x].son[1], val);    else return treap_rank(tr[x].son[0], val);}int main(){    int t;    while(~ scanf("%d", &t))    {        treap_init();        char ch;        int val, num = 0;        for(int i = 1; i <= t; i++)        {            scanf(" %c%d", &ch, &val);            if(ch == 'I')            {                if(! treap_find(treap_root, val)) treap_insert(treap_root, val), num++;            }            else if(ch == 'D')            {                if(treap_find(treap_root, val)) treap_del(treap_root, val), num--;            }            else if(ch == 'K')            {                if(num < val) printf("invalid\n");                else printf("%d\n", treap_kth(treap_root, val));            }            else printf("%d\n", treap_rank(treap_root, val));        }    }    return 0;}

pb_ds：

#include <bits/stdc++.h>#include <ext/pb_ds/assoc_container.hpp>#include <ext/pb_ds/tree_policy.hpp>using namespace std;using namespace __gnu_pbds;tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> bst;//用splay和ov_tree会TLE//tree<int, null_type, less<int>, splay_tree_tag, tree_order_statistics_node_update> bst;//tree<int, null_type, less<int>, ov_tree_tag, tree_order_statistics_node_update> bst;int main(){    int t, val;    char ch;    scanf("%d", &t);    for(int i = 1; i <= t; i++)    {        scanf(" %c%d", &ch, &val);        if(ch == 'I') bst.insert(val);        else if(ch == 'D') bst.erase(val);        else if(ch == 'K')        {            if(bst.size() >= val) printf("%d\n", *bst.find_by_order(val-1));//返回的是迭代器            else printf("invalid\n");        }        else printf("%d\n", bst.order_of_key(val));//统计比val小的值的个数，是严格小于    }    return 0;}

用树状数组或者线段树离线也可以做，核心思想就是维护前缀和。首先把数据离散化，对于插入操作，先检查树状数组里面有没有这个值，如果没有则往树状数组里面直接插入，值为1。对于删除操作，往树状数组里面插入，值为-1。对于第k大值，可以二分枚举答案，用枚举值的前缀和来判断当前枚举值是不是第k大。对于统计小于x的值的个数，直接求x-1的前缀和就好了。跑的比treap都快了一点。。。。

#include <bits/stdc++.h>using namespace std;const int N = 200000 + 10, INF = 0x3f3f3f3f;bool vis[N];struct BIT{    int n, b[N];    void init(int _n)    {        n = _n;        memset(b, 0, sizeof b);    }    void add(int i, int x)    {        for(; i <= n; i += i & -i) b[i] += x;    }    int sum(int i)    {        int ans = 0;        for(; i >= 1; i -= i & -i) ans += b[i];        return ans;    }}bit;char op[N];int a[N], b[N];int main(){    int t;    scanf("%d", &t);    int k = 0;    for(int i = 1; i <= t; i++)    {        scanf(" %c%d", &op[i], &a[i]);        b[++k] = a[i];    }    sort(b + 1, b + 1 + k);    k = unique(b + 1, b + 1 + k) - b - 1;    bit.init(k);    memset(vis, 0, sizeof vis);    int num = 0;    for(int i = 1; i <= t; i++)    {        int tmp = lower_bound(b + 1, b + 1 + k, a[i]) - b;        if(op[i] == 'I')        {            if(! vis[tmp]) bit.add(tmp, 1), num++, vis[tmp] = true;        }        else if(op[i] == 'D')        {            if(vis[tmp]) bit.add(tmp, -1), num--, vis[tmp] = false;        }        else if(op[i] == 'K')        {            if(num < a[i]) puts("invalid");            else            {                int l = 1, r = k, res = 0;                while(l <= r)                {                    int mid = (l + r) >> 1;                    if(bit.sum(mid) >= a[i]) res = b[mid], r = mid - 1;                    else l = mid + 1;                }                printf("%d\n", res);            }        }        else printf("%d\n", bit.sum(tmp - 1));    }    return 0;}