CF 1381B Unmerge（思惟 + 01背包肯定可行解）

题目：

Let a and b be two arrays of lengths n and m, respectively, with no elements in common. We can define a new array merge(a,b) of length n+m recursively as follows:

• If one of the arrays is empty, the result is the other array. That is, merge(∅,b)=b and merge(a,∅)=a. In particular, merge(∅,∅)=∅.
• If both arrays are non-empty, and a1<b1, then merge(a,b)=[a1]+merge([a2,…,an],b). That is, we delete the first element a1 of a, merge the remaining arrays, then add a1 to the beginning of the result.
• If both arrays are non-empty, and a1>b1, then merge(a,b)=[b1]+merge(a,[b2,…,bm]) That is, we delete the first element b1 of b, merge the remaining arrays, then add b1 to the beginning of the result.

This algorithm has the nice property that if a and b are sorted, then merge(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, if a=[3,1] and b=[2,4]，then merge(a,b)=[2,3,1,4].

A permutation is an array consisting of nn distinct integers from 11 to nn in 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=3 but there is 4 in the array).

There is a permutation p of length 2n. Determine if there exist two arrays a and b, each of length n and with no elements in common, so that p=merge(a,b).

 

思路：咱们容易想到“3 2”，“7 1”这些状况，这两个数必定属于同一个集合且是连续的，根据这个规律看“7 1 6”，咱们发现若是6和 7 1不属于同一个集合的话，那么6必定时另外一个集合的头元素，那么6必定不可能出如今7的后面，能够推出“7 1 6”属于同一个集合，这样咱们能够找到一个规律，例如：

6 1 3 7 4 5 8 2，咱们能够分红[6 1 3] [7 4 5] [8 2]；3 2 6 1 5 7 8 4,咱们能够分红[3 2] [6 1 5] [7] [8 4].这样咱们能够把每一个块的个数统计，而后咱们只须要肯定这些数字是否是能够组成n便可。

 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 }