10-排序6 Sort with Swap(0, i)（浙大数据结构PTA习题）

10-排序6 Sort with Swap(0, i)

分数 25        作者 陈越        单位 浙江大学

Question:

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

代码长度限制:16 KB        时间限制:400 ms        内存限制:64 MB

Answer:

示例推演：

发现与结论： 

代码展示： 

# include<stdio.h>

// 返回一个数组中大环的个数 
int CycleNum(int Array[], int N);

int main(){
	// 接收数组大小 
	int N;
	scanf("%d",&N);
	int Array[N];
	// 接收元素并统计 
	int i,Position,SmallCount=0;
	for(i=0;i<N;i++){
		scanf("%d",&Array[i]);
		if(Array[i]==0){
			// 存储开始时0元素的位置
			Position = i; 
		}
		if(Array[i] == i){
			// 统计一下自成环的个数，并将其位置标记为-1 
			Array[i] = -1;
			SmallCount++;
		}
	}
	// 检查一下0是否在位置0上（即是否自成环） 
	int valid = 0;
	if(Position == 0)valid = 1; 
	// 统计一下这个数组中大环的个数（至少包含2个元素才能称为大环）
	int BigCount = CycleNum(Array,N);
	// 计算所有大环中元素的总个数
	int BigNum = N - SmallCount; 
	// 计算最终交换次数 
	int Res = BigNum + BigCount;
	if(valid==0)Res -= 2;
	printf("%d",Res); 
	return 0; 
}

// 统计一个数组有多少个大环
int CycleNum(int Array[], int N){
	int i,j,tmp,Count = 0;
	for(i=0;i<N;i++){
		// 使用-1作为该位置元素已在某个环中的标志 
		if(Array[i]!=-1){
			while(Array[i]!=-1){
				tmp = i;
				i = Array[i]; 
				Array[tmp] = -1;
			}
			// 大环的数量加1
			Count++; 
		}
	}
	return Count;
}

运行结果：