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.
解答：
该题我刚开始始终不太理解，后来自己用笔模拟了一下过程有了比较清晰的思路。
即，用0和这个位置上本应该存在的数交换，如果0在过程中被换到了0位置上，而还存在其他没有换的数，那么将0和其他未换的随机一个数进行交换，使交换过程继续进行 ，使得最后序列有序。
但是，如果用一个数组存储初始序列，下标i对应输入的第i个数时，应该也可以解决问题，但是N <= 10^5让我们警惕，即可能出现超时情况，因此我们可以利用输入数的唯一性，将输入数作为下标，数组中存储位置信息i，这样便省去了O(n)查找的麻烦。
AC代码如下：
#include<iostream>
#include<cstdio>
#include<vector>using namespace std;int main()
{int n, val, ans = 0, count = 0, index = 1;scanf("%d", &n);vector<int> loc(n);for(int i = 0; i < n; i++){scanf("%d", &val);loc[val] = i;if(val != i && val != 0) count++;}while(count){if(loc[0] == 0){while(index < n){if(loc[index] != index){swap(loc[index], loc[0]);ans++;break;}index++;}}while(loc[0] != 0){swap(loc[loc[0]], loc[0]);ans++;count--;}}printf("%d\n", ans);return 0;
}