1067. Sort with Swap(0,*)

1067. Sort with Swap(0,*) (25)

时间限制

150 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

9

#include<stdio.h>#include<vector>using namespace std;int main(){//freopen("F://Temp/input.txt", "r", stdin);int n;scanf("%d", &n);vector<int> v;for(int i = 0; i < n; i ++){int x;scanf("%d", &x);v.push_back(x);}int count  = 0;bool flag = 0;//0是不是在正确位置上 if(v[0] == 0)flag = true;int num = 0;//组数目for(int i = 0; i < n; i ++){bool newGroup = false;while(v[i] != i){newGroup = true;int tmp = v[v[i]];v[v[i]] = v[i];v[i] = tmp;count ++;}if(newGroup)num ++;}if(flag)count += num*2;elsecount += (num-1)*2;printf("%d\n", count);return 0;}