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.
记录下每个数的位置，例如pos[4]=0表示在输入序列中4是第0个数。假设pos[0]==i，当i不等于0时，表示数字0占用了数字i的位置，这时，把数字i换到第i个位置上，使其处于正确位置，而把0换到数字i原来的位置上；如果pos[0]==0，就找到一个不在正确位置上的数字，假设为k，将0与k的位置交换，重复pos[0]!=0时的步骤。比较像贪心，走一步看一步。
为了避免超时，需要做如下处理：当pos[0]==0，在寻找不在正确位置上的第一个数时，不要从头（1）开始，要设置一个标记k，记录的是上一次找到的不在正确位置上的第一个数，这次寻找就从k开始。
#include<iostream>
using namespace std;
const int maxn = 100010;
int pos[maxn];
void swap(int& a, int& b) {
int temp = a;a = b;b = temp;
}
int main() {
int n, wrongPos, ans = 0, k = 1;cin >> n;wrongPos = n;//不在正确位置上的数的个数for (int i = 0; i < n; i++) {
int d;cin >> d;pos[d] = i;if (d == i && d != 0)wrongPos--;}while (wrongPos != 1) {
//0的位置不需要考虑，当其他数位置都正确时，0的位置自然也正确if (pos[0] == 0) {
while (k < n && pos[k] == k)k++;swap(pos[0], pos[k]);ans++;}swap(pos[0], pos[pos[0]]);wrongPos--;ans++;}cout << ans << endl;return 0;
}