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

 

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

 

一开始按照选择排序做了， 结果有两个测试点超时， 看看网上的思路基本上都是判断是否被访问的算法，想了想这个应该是表排序的变形，就按照表排序方式做了，花了一下午调试 果然AC了， 特地来交流 ^_^!

 

1 #include 
     
       2 #include 
      
        3  4 typedef int ElementType; 5  6 void Swap(ElementType *a, ElementType *b); 7 void PrintA(ElementType A[], int N); 8 int IsSort( ElementType A[], int m, int N); 9 void SwapZero( ElementType A[], ElementType B[], int N);10 11 typedef struct {12     int index;//0元素所在下标13     int count;//记录交换次数14 }Zero;15 16 Zero zero;17 18 int main(){19     int N, i;20     zero.count = 0;21     //freopen("C:\\in.txt","r", stdin);22     scanf("%d", &N);23     ElementType* A;//这个是原来的数组24     A = (ElementType*)malloc(N*sizeof(ElementType));25     ElementType* B;//这个是表26     B = (ElementType*)malloc(N*sizeof(ElementType));27     for( i=0; i