1067 Sort with Swap(0, i) （25 分）思路

2024-04-15 09:38:34

1067 Sort with Swap(0, i) （25 分）

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:

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 (≤105) 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:

10
3 5 7 2 6 4 9 0 8 1

Sample Output:

经过演算发现有 0 的环是 n-1 没0的环是n+1当然了个数要大于等于2才行

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int digit[maxn], pre[maxn];
int cnt[maxn];
int find(int x){if(x == pre[x])return x;return pre[x] = find(pre[x]);
}
void merge(int x, int y)
{int fx = find(x);int fy = find(y);if(fx != fy){pre[fx] = fy;}
}
int main()
{ios::sync_with_stdio(false);int n;cin>>n;for(int i = 0; i < n ; i++){cin>>digit[i];  pre[i] = i;}   for(int i = 0 ; i < n; i ++)merge(i , digit[i]);for(int i = 0; i < n; i ++)cnt[find(i)] ++;   int gg = cnt[find(0)];int kk = find(0);int ans = 0;if(gg != 1) //表明不是digit[0] = 0 ans = gg - 1; for(int i = 0 ; i < n ; i++){if(cnt[i] == 1 || i == kk || cnt[i] == 0)continue;ans = ans + cnt[i] + 1; } cout<<ans<<endl;return 0;
}

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