1067 Sort with Swap(0, i) (25分)

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

题意：给你n个数的序列（0~n-1），交换排序的时候只能跟0交换，当序列有序的时候求最少的交换次数

方法：贪心：已经在合适位置上的元素就不需要动了，我们需要记录除了0不在合适位置上的元素（代码中的cnt）；

看0所在的位置，如果0不在位置0，那么0在哪个位置就和那个元素交换；如果0在位置0，因为我们需要借助0使其他的元素归位，所以就找第一个不在适合位置上的元素与0交换

#include <bits/stdc++.h>
#define Max 110005
using namespace std;
int n, a[Max];
int pos[Max];
int cnt, ans; 
int main(){
  memset(a, 0, sizeof(a));
  memset(pos, 0, sizeof(pos));
  ans = 0;
  cin >> n;
  cnt = n - 1;
  for(int i = 0; i < n; i++){
    cin >> a[i];
    pos[a[i]] = i;
    if(a[i] == i && a[i] != 0) cnt--;//需要交换的 
  }
  int j = 1;
  while(cnt){
    if(pos[0] != 0){
      swap(pos[0], pos[pos[0]]);
      cnt--;
      ans++;
    }else{
      for(;j < n; j++){  //前边的顺序都安排好了 每次从 j 开始遍历 如果 每次 从头遍历 会超时 
        if(pos[j] != j){
          swap(pos[0], pos[j]);
          ans++;
          break;
        }
      }
    }
  } 
  cout << ans << endl;
  return 0;
}