题目
中位数
多次询问,每次修改数组中一个数,问修改后n个数的中位数
思路
- 使用
multiset
,分别维护数组的较大的\(n/2+1\)个和较小的\(n/2\)个; - 根据数据范围,或许可用线段树+二分...
代码
Code
#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
#include <set>
using namespace std;
using LL = long long;
const int N = 1e6 + 10;
int a[N];
int b[N];
void print(multiset<int> &a)
{
for (auto it: a) cout << it << ' ';
cout << endl;
}
void solv()
{
int n, m;
cin >> n >> m;
for (int i = 1; i <= n; i ++)
{
cin >> a[i];
b[i] = a[i];
}
sort(b+1, b+1+n);
multiset<int> big, small;
int mid = n/2 + 1;
for (int i = 1; i < mid; i ++) small.insert(b[i]);
for (int i = mid; i <= n; i ++) big.insert(b[i]); // mid -> big.begin
int p, x;
for (int i = 1; i <= m; i ++)
{
// cout << "i = " << i << endl;
cin >> p >> x;
auto pos = small.find(a[p]);
if (pos != small.end()) small.erase(pos);
else big.erase(big.find(a[p]));
big.insert(x); a[p] = x;
while (big.size() > small.size() + 1)
{
auto pos1 = big.begin();
small.insert(*pos1);
big.erase(pos1);
}
auto pos1 = small.end(); pos1 --;
auto pos2 = big.begin();
while(*pos1 > *pos2)
{
big.insert(*pos1); small.insert(*pos2);
big.erase(pos2); small.erase(pos1);
pos1 = small.end(); pos1 --;
pos2 = big.begin();
}
cout << (*pos2) << endl;
// cout << "small: " ; print(small);
// cout << "big: " ; print(big);
}
}
int main()
{
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int T = 1;
// cin >> T;
while (T --)
{
solv();
}
return 0;
}