CodeForces 1382D Unmerge （思维，背包）

题目链接：点击这里

题目大意：

题目分析：
发现，对于每一个 $x$ 若后面是 $y,z$ 且 $x>y,x>z$ 那么 $x$ 和 $y,z$ 一定是同一组的，然后利用这个性质将序列划分，再对划分好的数组长度跑一下背包看其是否可以恰好组成两个长度为 $n$ 的序列即可

具体细节见代码：

// Problem: B - Unmerge
// Contest: Virtual Judge - 康复训练01
// URL: https://vjudge.net/contest/485494#problem/B
// Memory Limit: 1024 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

//#pragma GCC optimize(2)
//#pragma GCC optimize("Ofast","inline","-ffast-math")
//#pragma GCC target("avx,sse2,sse3,sse4,mmx")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<unordered_map>
#define ll long long
#define inf 0x3f3f3f3f
#define Inf 0x3f3f3f3f3f3f3f3f
//#define int  ll
#define endl '\n'
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
using namespace std;
int read()
{
	int res = 0,flag = 1;
	char ch = getchar();
	while(ch<'0' || ch>'9')
	{
		if(ch == '-') flag = -1;
		ch = getchar();
	}
	while(ch>='0' && ch<='9')
	{
		res = (res<<3)+(res<<1)+(ch^48);//res*10+ch-'0';
		ch = getchar();
	}
	return res*flag;
}
const int maxn = 1e6+5;
const int mod = 1e9+7;
const double pi = acos(-1);
const double eps = 1e-8;
int n,a[maxn],dp[maxn];
vector<int>v;
int main()
{
	int t = read();
	while(t--)
	{
		n = read();
		for(int i = 1;i <= 2*n;i++) a[i] = read();
		v.clear();
		for(int i = 1,r;i <= 2*n;i = r+1)
		{
			r = i+1;
			while(r<=2*n && a[i]>a[r]) r++;
			r--;
			v.push_back(r-i+1);
		}
		for(int i = 1;i <= n;i++) dp[i] = 0;
		dp[0] = 1;
		for(auto it:v)
			for(int i = n;i >= it;i--) dp[i] |= dp[i-it];
		puts(dp[n] ? "YES" : "NO");
	}
	return 0;
}