CF 1381B Unmerge（思维 + 01背包确定可行解）

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：我们容易想到“3 2”，“7 1”这些情况，这两个数一定属于同一个集合且是连续的，根据这个规律看“7 1 6”，我们发现如果6和 7 1不属于同一个集合的话，那么6一定时另一个集合的头元素，那么6一定不可能出现在7的后面，可以推出“7 1 6”属于同一个集合，这样我们可以找到一个规律，例如：

6 1 3 7 4 5 8 2，我们可以分成[6 1 3] [7 4 5] [8 2]；3 2 6 1 5 7 8 4,我们可以分成[3 2] [6 1 5] [7] [8 4].这样我们可以把每个块的个数统计，然后我们只需要确定这些数字是不是可以组成n即可。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <string>
#include <vector>
#include <cmath>
 
using namespace std;
 
#define ll long long
#define pb push_back
#define fi first
#define se second
 
const int N = 2e5 + 10;
bool f[N];
int a[N];
 
void solve()
{      
    int T;
    cin >> T;
    while(T--){
        int n;
        cin >> n;
        n <<= 1;
        for(int i = 1; i <= n; ++i) cin >> a[i];
        // for(int i = 1; i <= n; ++i) cout << a[i] << " ";
        // cout << endl;
        vector<int > v;
        int x = a[1];
        int cnt = 0, inx = 1;
        while(1){
            if(a[inx] <= x) cnt++, inx++;
            else{
                v.pb(cnt);
                cnt = 0;
                x = a[inx];
            }
            if(inx > n) break;
        }
        if(cnt > 0) v.pb(cnt);
       // cout << "n = " << n << endl;
        n >>= 1;
        for(int i = 0; i <= n; ++i) f[i] = false;
        f[0] = true;
        //cout << " f[n] = " << f[n] << endl;
        for(auto vv : v){
            for(int i = n; i >= 0; --i){
                if(i - vv < 0) break;
                if(f[i - vv] == true) f[i] = true;
            }
        }
        //cout << "n = " << n << endl;
        if(f[n] == true) cout << "YES" << endl;
        else cout << "NO" << endl;
    }
}
 
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0); 
    solve();
 
    return 0;
}