1067. Sort with Swap(0, i)

Posted on | In , |

题目

https://pintia.cn/problem-sets/994805342720868352/problems/994805403651522560

Given any permutation of the numbers {0, 1, 2,…, N−1}, it is easy to sort them in increasing order. But what if Swap(0, *) is the ONLY operation that is allowed to use? For example, to sort {4, 0, 2, 1, 3} we may apply the swap operations in the following way:

1
2
3
Swap(0, 1) => {4, 1, 2, 0, 3}
Swap(0, 3) => {4, 1, 2, 3, 0}
Swap(0, 4) => {0, 1, 2, 3, 4}

Now you are asked to find the minimum number of swaps need to sort the given permutation of the first N nonnegative integers.

Input Specification:
Each input file contains one test case, which gives a positive N (≤10^5​​) followed by a permutation sequence of {0, 1, …, N−1}. All the numbers in a line are separated by a space.

Output Specification:
For each case, simply print in a line the minimum number of swaps need to sort the given permutation.

Sample Input:

1
2
10
3 5 7 2 6 4 9 0 8 1

Sample Output:

1
9

想法

一眼看过去，应该是一道技巧题，不是常规的数据结构和算法。

每一步，把逆序数降到最低。
好像不止，要把每一个都调整到对应的位置。

唉，那是不是，多少个不在自己位置的，就换多少次对了。

怎么有点连通分量的味道，如果哪几个之间错位了，就需要借助一次0位置进行调整。

如果k个之间错位：
包括0：k-1次；不包括0：k+1次

唉，AC了，果然是trick啊，好好想，细细想，这种题估计就是一个小转弯，工作量并不大

答案

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
//
// Created by Zhao Xiaodong on 2018/8/23.
//
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int N;
    cin >> N;
    if (N == 0) {
        cout << 0;
        return 0;
    }
    vector<int> a(N);
    for (int i = 0; i < N; i++) {
        cin >> a[i];
    }
    vector<int> marked(N, false);
    int res = 0;
    if (a[0] != 0) {
        marked[0] = true;
        for (int i = a[0]; i != 0; i = a[i]) {
            marked[i] = true;
            res++;
        }
    }
    for (int i = 1; i < N; i++) {
        if (a[i] != i && !marked[i]) {
            res += 2;
            marked[i] = true;
            for (int j = a[i]; j != i && !marked[j]; j = a[j]) {
                marked[j] = true;
                res++;
            }
        }
    }
    cout << res;
    return 0;
}