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 (≤10​5​​) 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

解題思路

基本思路是,我們依次拿0與0所在位置代表的那個數(比如測試用例中0的位置為7）進行交換,最後得到有序序列

不過這樣計算會出現一種特殊情況:就是0和0位置那個數發生了交換,那麼演算法到這裡就終止了

遇到這種特殊情況,只需將0與最小的那個不在本位上的數交換就可以了。

也就是說

i != m[i]時,將0與數交換, i = m[i],繼續遍歷,程式碼如下:

#include <iostream>
#include <algorithm>
#include <string.h>
#include <string>
using namespace std;
const int MAXN = 100010;
int Numbers[MAXN];
int HashTable[MAXN];  //用來儲存一個數所在的位置數
int main() {
	int N;
	scanf("%d", &N);
	for (int i = 0; i < N; ++i) {
		scanf("%d", &Numbers[i]);  //依次讀入
		HashTable[Numbers[i]] = i;
	}
	int pos = HashTable[0];  //給出0的位置
	int counter = 0;
	//中斷迴圈條件:當每個i == m[i]時
	while (true) {
		int tempPos = pos;
		int i;
		if (tempPos == 0) {
			//遍歷尋找一個 i != m[i]的數字
			for (i = 1; i < N; ++i) {
				if (i != Numbers[i]) {
					swap(Numbers[tempPos], Numbers[i]);
					swap(HashTable[Numbers[tempPos]], HashTable[Numbers[i]]);
					counter++;
					pos = i;
					break;
				}
			}
			if (i == N) {
				break;   //陣列已有序 退出迴圈即可
			}
		}
		else {
			pos = HashTable[pos];
			swap(Numbers[pos], Numbers[tempPos]);
			swap(HashTable[Numbers[pos]], HashTable[Numbers[tempPos]]);
			counter++;
		}
	}
	printf("%d", counter);
	system("PAUSE");
	return 0;
}