PAT 1067. Sort with Swap(0,*)
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 1Sample 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 ^_^!
#include <stdio.h> #include <stdlib.h> typedef int ElementType; void Swap(ElementType *a, ElementType *b); void PrintA(ElementType A[], int N); int IsSort( ElementType A[], int m, int N); void SwapZero( ElementType A[], ElementType B[], int N); typedef struct { int index;//0元素所在下标 int count;//记录交换次数 }Zero; Zero zero; int main(){ int N, i; zero.count = 0; //freopen("C:\\in.txt","r", stdin); scanf("%d", &N); ElementType* A;//这个是原来的数组 A = (ElementType*)malloc(N*sizeof(ElementType)); ElementType* B;//这个是表 B = (ElementType*)malloc(N*sizeof(ElementType)); for( i=0; i<N; i++){ scanf("%d", &A[i]); B[A[i]] = i;//记录A[i]所在的位置 if(A[i] == 0) zero.index = i; } SwapZero(A, B, N); printf("%d\n", zero.count); return 0; } int IsSort( ElementType A[], int m, int N){ int i; int flag = 0; for(; m<N; m++ ){ if( m != A[m] ) { flag = m; break; } } return flag; } void SwapZero( ElementType A[], ElementType B[], int N){ int i, m = 1; for( ; ; ){ if( zero.index != 0 ){//交换swap(0,i); Swap( &A[zero.index], &A[B[zero.index]]); B[0] = B[zero.index]; B[A[zero.index]] = zero.index;//更新表 zero.index = B[0]; zero.count++; } else if( m=IsSort(A, m, N)){ //找到第一个位置不对的数字交换 Swap( &A[zero.index], &A[m]);//交换,更新表 B[zero.index] = m; B[A[zero.index]] = 0; zero.index = B[zero.index]; zero.count++; } else break; } } void Swap(ElementType *a, ElementType *b){ int tmp; tmp = *a; *a = *b; *b = tmp; }
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