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CF 1427D Unshuffling a Deck

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题目描述

https://codeforces.com/problemset/problem/1427/D

Solution

如果序列没有排好序，那么一定存在 $i<j,a_i=a_j+1$

那么我们考虑将序列分成四部分 $[1,i-1],[i,k],[k+1,j],[j+1,n]$

两边有可能是空的，但是并没有影响，注意中间的 $k$ 的作用是不让已经连续的子段分开

也就是我们要找一个 $i\le k< j, a_k>a_{k+1}$，这样每次都使连续的段加 $1$

最多需要做 $n-1$ 次

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#include <iostream>
#include <cstdio>
#include <vector>
#define pb push_back
#define maxn 53
using namespace std;

int n, a[maxn], b[maxn], A[maxn];

vector<int> ans[maxn]; int c1; 
int main() {
    cin >> n; a[0] = 0; 
    for (int i = 1; i <= n; ++i) cin >> a[i], A[a[i]] = i; 
    while (1) {
        int l = -1, r, k; 
        for (int i = 1; i < n; ++i)
            if (A[a[i] - 1] > i) l = i, r = A[a[i] - 1]; 
        for (int i = l; i < r; ++i)
            if (a[i] > a[i + 1]) k = i;
        if (l == -1) break; ++c1;
        
        int s1 = l - 1, s2 = k - l + 1, s3 = r - k, s4 = n - r; 
        for (int i = 1; i <= n; ++i) b[i] = a[i];
        for (int i = 1; i <= s4; ++i) a[i] = b[r + i];
        for (int i = 1; i <= s3; ++i) a[s4 + i] = b[k + i];
        for (int i = 1; i <= s2; ++i) a[s4 + s3 + i] = b[l + i - 1];
        for (int i = 1; i <= s1; ++i) a[s4 + s3 + s2 + i] = b[i];
        for (int i = 1; i <= n; ++i) A[a[i]] = i; 
        ans[c1].pb(2 + (s1 > 0) + (s4 > 0));
        ans[c1].pb(s1); ans[c1].pb(s2); ans[c1].pb(s3); ans[c1].pb(s4);
    }
    cout << c1 << endl;
    for (int i = 1; i <= c1; ++i, puts(""))
        for (auto u : ans[i]) if (u) cout << u << " "; 
    return 0;
}