树状数组 ---- 树状数组+动态维护前缀中位数 D. Omkar and Medians

题目大意：

解题思路：

首先我们看他们的定义：
bib_ibi是a1,a2,a3,....,ai,ai+1,.......a2i−1a_1,a_2,a_3,....,a_i,a_{i+1},.......a_{2i-1}a1,a2,a3,....,ai,ai+1,.......a2i−1
bi+1b_{i+1}bi+1是a1,a2,a3,....,ai,ai+1,.......a2i+1a_1,a_2,a_3,....,a_i,a_{i+1},.......a_{2i+1}a1,a2,a3,....,ai,ai+1,.......a2i+1
我们发现bi+1b_{i+1}bi+1比bib_ibi只多了a2i,a2i+1a_{2i},a_{2i+1}a2i,a2i+1
那么就每次加入两个数！

那么我们看一下每一次中位数会怎么变换？
假设c1,c2,c3.....ci,ci+1,......c2i−1c_1,c_2,c_3.....c_i,c_{i+1},......c_{2i-1}c1,c2,c3.....ci,ci+1,......c2i−1是a1,a2,a3,....,ai,ai+1,.......a2i−1a_1,a_2,a_3,....,a_i,a_{i+1},.......a_{2i-1}a1,a2,a3,....,ai,ai+1,.......a2i−1排好序的结果
那么bi=ci中位数b_i=c_i中位数bi=ci中位数

对于a2i,a2i+1a_{2i},a_{2i+1}a2i,a2i+1有三种情况

它们分别是在cic_ici两端那么中位数不变
它们有两个都在左边,那么中位数会往右边移动一位！
它们有两个都在右边,那么中位数会往左边移动一位！

那么对于相邻的两个中位数bi和bi+1b_i和b_{i+1}bi和bi+1是没有数在它们之间，因为每次只移动一位！
这个可以用树状数组去维护，不过要先离散化

#include <bits/stdc++.h>
#define mid ((l + r) >> 1)
#define Lson rt << 1, l , mid
#define Rson rt << 1|1, mid + 1, r
#define ms(a,al) memset(a,al,sizeof(a))
#define log2(a) log(a)/log(2)
#define lowbit(x) ((-x) & x)
#define IOS std::ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define INF 0x3f3f3f3f
#define LLF 0x3f3f3f3f3f3f3f3f
#define f first
#define s second
#define endl '\n'
using namespace std;
const int N = 2e6 + 10, mod = 1e9 + 9;
const int maxn = 500010;
const long double eps = 1e-5;
const int EPS = 500 * 500;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll,ll> PLL;
typedef pair<double,double> PDD;
template<typename T> void read(T &x)
{x = 0;char ch = getchar();ll f = 1;while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
template<typename T, typename... Args> void read(T &first, Args& ... args)
{read(first);read(args...);
}
int tr[maxn];
int arr[maxn];
vector<int> lis;
map<int,int> mp;inline void add(int pos, int val) {while(pos < maxn) {tr[pos] += val;pos += lowbit(pos);}
}inline int sum(int x) {int res = 0;while(x) {res += tr[x];x -=lowbit(x);}return res;
}int main() {IOS;int T;cin >> T;while(T --) {int n;cin >> n;for(int i = 1; i <= n; ++ i) {cin >> arr[i];lis.push_back(arr[i]);}sort(lis.begin(),lis.end());lis.erase(unique(lis.begin(),lis.end()),lis.end());for(int i = 1; i <= n; ++ i) arr[i] = lower_bound(lis.begin(),lis.end(),arr[i]) - lis.begin() + 1;bool flag = 1;for(int i = 1; i <= n; ++ i) {if(!mp.count(arr[i])) {mp[arr[i]] = true;add(arr[i],1);}if(i == 1) continue;int x = sum(arr[i]);int y = sum(arr[i-1]);if(max(x,y) - min(x,y) > 1) {flag = 0;break;}}if(flag) cout << "YES\n";else cout << "NO\n";for(auto it : mp) add(it.first,-1);mp.clear();lis.clear();}
}