这个题本来是学树堆Treap练习的，但是我的板子超时，无奈之下去墙那边看了下，老外用了树状数组给解出来了，很巧妙的离散化操作。。。Orz。。。
ORDERSET - Order statistic set
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| English | Vietnamese |
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In this problem, you have to maintain a dynamic set of numbers which support the two fundamental operations
- INSERT(S,x): if x is not in S, insert x into S
- DELETE(S,x): if x is in S, delete x from S
and the two type of queries
- K-TH(S) : return the k-th smallest element of S
- COUNT(S,x): return the number of elements of S smaller than x
Input
- Line 1: Q (1 ≤ Q ≤ 200000), the number of operations
- In the next Q lines, the first token of each line is a character I, D, K or C meaning that the corresponding operation is INSERT, DELETE, K-TH or COUNT, respectively, following by a whitespace and an integer which is the parameter for that operation.
If the parameter is a value x, it is guaranteed that 0 ≤ |x| ≤ 109. If the parameter is an index k, it is guaranteed that 1 ≤ k ≤ 109.
Output
For each query, print the corresponding result in a single line. In particular, for the queries K-TH, if k is larger than the number of elements in S, print the word 'invalid'.
Example
Input
8
I -1
I -1
I 2
C 0
K 2
D -1
K 1
K 2
Output
1
2
2
invalid
#define _CRT_SECURE_NO_WARNINGS
#include <cstdio>
#include <algorithm>
using namespace std;
#define MAX 200005
#define pb push_back
int tree[200009], Hash[200009], save[200009];
inline int lowbit(int x) {
return x & -x;
}
void update(int pos, int val) {
for (int i = pos; i <= MAX; i += lowbit(i))
tree[i] += val;
}
int sum(int pos) {
int ans = 0;
for (int i = pos; i > 0; i -= lowbit(i))
ans += tree[i];
return ans;
}
struct Data {
int pos, val, Hash;
char type;
};
Data d[200009], query[200009];
bool cmp(const Data &a, const Data &b) {
return a.val == b.val ? a.pos < b.pos : a.val < b.val; //数据优先
}
int b_search(int k) {
int low = 0, high = 200000, mid;
while (low < high) {
mid = (low + high) >> 1;
if (sum(mid) >= k)
high = mid;
else
low = mid + 1;
}
return low;
}
int main() {
int q;
scanf("%d", &q);
for (int i = 0; i < q; i++) {
scanf(" %c%d", &d[i].type, &d[i].val);
d[i].pos = i + 1;
}
sort(d, d + q, cmp);
int cnt = 1, prev = d[0].val;
d[0].Hash = cnt;
save[cnt] = d[0].val;
for (int i = 1; i < q; i++) { //hash，不然不能用树状数组
if (prev != d[i].val) {
prev = d[i].val;
cnt++;
d[i].Hash = cnt;
save[cnt] = d[i].val;
}
else
d[i].Hash = cnt;
}
for (int i = 0; i < q; i++) {
query[d[i].pos] = d[i];
}
cnt = 0;
for (int i = 1; i <= q; i++) {
if (query[i].type == 'I') {
if (!Hash[query[i].Hash]) {
cnt++;
Hash[query[i].Hash] = 1;
update(query[i].Hash, 1);
}
}
else if (query[i].type == 'D') {
if (Hash[query[i].Hash]) {
cnt--;
Hash[query[i].Hash] = 0;
update(query[i].Hash, -1);
}
}
else if (query[i].type == 'K') {
if (query[i].val > cnt)
printf("invalid\n");
else
printf("%d\n", save[b_search(query[i].val)]);
}
else
printf("%d\n", sum(query[i].Hash - 1)); //查询小于query[i].hash的值
}
return 0;
}
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他的另一种解法（这也太神了吧）。现在不是大部分OJ支持这种操作（拓展的pb_ds, codeforces spoj 支持）
#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;
typedef tree<
int ,
null_type,
less<int>,
rb_tree_tag,
tree_order_statistics_node_update>pbd_set;
int main(){
pbd_set s;
int t;
scanf("%d ",&t);
while(t--){
char typ;
int n;
scanf(" %c%d",&typ , &n);
if(typ == 'I')
s.insert(n);
else if (typ == 'D')
s.erase(n);
else if (typ == 'K'){
int ans;
n--;
if(s.find_by_order(n) != s.end()){
ans = *s.find_by_order(n);
printf("%d\n",ans);
}
else
printf("invalid\n");
}
else{
int ans = s.order_of_key(n);
printf("%d\n",ans);
}
}
return 0;
}
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