題目：

Letaandbbe two arrays of lengthsnandm, respectively, with no elements in common. We can define a new arraymerge(a,b)of lengthn+mrecursively as follows:

• If one of the arrays is empty, the result is the other array. That is,merge(∅,b)=bandmerge(a,∅)=a. In particular,merge(∅,∅)=∅.
• If both arrays are non-empty, anda1<b1, thenmerge(a,b)=[a1]+merge([a2,…,an],b). That is, we delete the first elementa1ofa, merge the remaining arrays, then adda1to the beginning of the result.
• If both arrays arenon-empty, anda1>b1, thenmerge(a,b)=[b1]+merge(a,[b2,…,bm])That is, we delete the first elementb1ofb, merge the remaining arrays, then addb1to the beginning of the result.

This algorithm has the nice property that ifaandbare sorted, thenmerge(a,b)will also be sorted. For example, it is used as a subroutine in merge-sort. For this problem, however, we will consider the same procedure acting on non-sorted arrays as well. For example, ifa=[3,1]andb=[2,4]，thenmerge(a,b)=[2,3,1,4].

A permutation is an array consisting ofnndistinct integers from11tonnin arbitrary order. For example,[2,3,1,5,4]is a permutation, but[1,2,2]is not a permutation (2 appears twice in the array) and[1,3,4]is also not a permutation (n=3but there is4in the array).

There is a permutationpof length2n. Determine if there exist two arraysaandb, each of lengthnand with no elements in common, so thatp=merge(a,b).

思路：我們容易想到“3 2”，“7 1”這些情況，這兩個數一定屬於同一個集合且是連續的，根據這個規律看“7 1 6”，我們發現如果6和 7 1不屬於同一個集合的話，那麼6一定時另一個集合的頭元素，那麼6一定不可能出現在7的後面，可以推出“7 1 6”屬於同一個集合，這樣我們可以找到一個規律，例如：

6 1 3 7 4 5 8 2，我們可以分成[6 1 3] [7 4 5] [8 2]；3 2 6 1 5 7 8 4,我們可以分成[3 2] [6 1 5] [7] [8 4].這樣我們可以把每個塊的個數統計，然後我們只需要確定這些數字是不是可以組成n即可。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <queue>
 5 #include <string>
 6 #include <vector>
 7 #include <cmath>
 8  
 9 using namespace std;
10  
11 #define ll long long
12 #define pb push_back
13 #define fi first
14 #define se second
15  
16 const int N = 2e5 + 10;
17 bool f[N];
18 int a[N];
19  
20 void solve()
21 {      
22     int T;
23     cin >> T;
24     while(T--){
25         int n;
26         cin >> n;
27         n <<= 1;
28         for(int i = 1; i <= n; ++i) cin >> a[i];
29         // for(int i = 1; i <= n; ++i) cout << a[i] << " ";
30         // cout << endl;
31         vector<int > v;
32         int x = a[1];
33         int cnt = 0, inx = 1;
34         while(1){
35             if(a[inx] <= x) cnt++, inx++;
36             else{
37                 v.pb(cnt);
38                 cnt = 0;
39                 x = a[inx];
40             }
41             if(inx > n) break;
42         }
43         if(cnt > 0) v.pb(cnt);
44        // cout << "n = " << n << endl;
45         n >>= 1;
46         for(int i = 0; i <= n; ++i) f[i] = false;
47         f[0] = true;
48         //cout << " f[n] = " << f[n] << endl;
49         for(auto vv : v){
50             for(int i = n; i >= 0; --i){
51                 if(i - vv < 0) break;
52                 if(f[i - vv] == true) f[i] = true;
53             }
54         }
55         //cout << "n = " << n << endl;
56         if(f[n] == true) cout << "YES" << endl;
57         else cout << "NO" << endl;
58     }
59 }
60  
61 int main()
62 {
63     ios::sync_with_stdio(false);
64     cin.tie(0);
65     cout.tie(0); 
66     solve();
67  
68     return 0;
69 }