PAT甲级1067. Sort with Swap(0,*) (25)

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本文探讨了一种特殊情况下的序列排序问题：仅允许使用Swap(0, *)操作来对{0,1,2,...,N-1}

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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

将各个数存入数组P[];
如果0在pos处(pos!=0)，即P[pos]=0，就交换0和pos，如题目所给测试用例，第一步0和7交换，这样7就归位了，然后0到了pos=2处，再交换0和2；
如果在交换过程中，0到了pos=0怎么办呢？就让0和随便什么还没有归位的数交换，这样又可以继续执行之前描述的过程。

Customized Test Case
Input
4 0 1 3 2
Output
3

#include <cstdio>
using namespace std;
#include <vector>
#include <algorithm>
#include <map>

const int MAX=100000+10;
int P[MAX];
map<int,int> M; //first:数字; second:在P数组中的下标

int main(){
  int N;
  scanf("%d",&N);

  int pos0=-1;//0的位置
  int count=0;//已经在正确位置上的数字。0不算
  int num=0;//交换的次数
  for(int i=0;i<N;i++) {
    scanf("%d",&P[i]);
    if(P[i]==i) {
      if(P[i]!=0) count++;
    }
    else{
      if(P[i]!=0) M[P[i]]=i;
    }
    if(P[i]==0) pos0=i;
  }

  int tmp;
  while(count<N-1){
  //  printf("pos0=%d\n",pos0);
    if(pos0!=0) {
      count++;
      num++;
      tmp=M[pos0];
      M.erase(pos0);
    //  printf("swap1 %d %d\n",P[pos0],P[tmp]);
      swap(P[tmp],P[pos0]);
      pos0=tmp;
    }
    else{
        num++;
        map<int,int>::iterator it=M.begin();
        tmp=it->second;
    //    printf("%d %d\n",it->first,it->second);
        M[it->first]=pos0;
      //  printf("swap2 %d %d\n",P[pos0],P[tmp]);
        swap(P[pos0],P[tmp]);
        pos0=tmp;
    }
  }

  printf("%d",num);

  return 0;
}