PAT 1067 Sort with Swap(0, i) (贪心法)

梦里梦外; 2022-04-10 14:41 126阅读 0赞

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: ###

解题思路

基本思路是,我们依次拿0与0所在位置代表的那个数(比如测试用例中0的位置为7）进行交换,最后得到有序序列

不过这样计算会出现一种特殊情况:就是0和0位置那个数发生了交换,那么算法到这里就终止了

遇到这种特殊情况,只需将0与最小的那个不在本位上的数交换就可以了。

也就是说

i != m\[i\]时,将0与数交换, i = m\[i\],继续遍历,代码如下:

#include <iostream>
    #include <algorithm>
    #include <string.h>
    #include <string>
    using namespace std;
    const int MAXN = 100010;
    int Numbers[MAXN];
    int HashTable[MAXN];  //用来保存一个数所在的位置数
    int main() {
    	int N;
    	scanf("%d", &N);
    	for (int i = 0; i < N; ++i) {
    		scanf("%d", &Numbers[i]);  //依次读入
    		HashTable[Numbers[i]] = i;
    	}
    	int pos = HashTable[0];  //给出0的位置
    	int counter = 0;
    	//中断循环条件:当每个i == m[i]时
    	while (true) {
    		int tempPos = pos;
    		int i;
    		if (tempPos == 0) {
    			//遍历寻找一个 i != m[i]的数字
    			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;   //数组已有序 退出循环即可
    			}
    		}
    		else {
    			pos = HashTable[pos];
    			swap(Numbers[pos], Numbers[tempPos]);
    			swap(HashTable[Numbers[pos]], HashTable[Numbers[tempPos]]);
    			counter++;
    		}
    	}
    	printf("%d", counter);
    	system("PAUSE");
    	return 0;
    }