7-16 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 (≤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

这道题用到了表排序中N个数字的排列由若干独立的环组成的知识。

1.单元环swap(0,i)次数为0；

2.有0的非单元环，swap（0，i)次数为n-1;

3.无0的单元环，可以先把环里随便一个数和0交换一次，这样整个环就变成了含0的n+1个元素的非单元环，根据前面的情况2，次数为(n+1)-1,但还要加上把0换进去的那次，故swap（0，i)次数为n+1;

代码如下：

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int flag=0; // 记录circle是否含0；
int main(){int N; cin>>N;// 储存数列vector<int> sequence; // 为了已知元素找下标而建立的position集合// 其中pos[i]表示sequence中数值为i的元素的下标为pos[i]vector<int> pos;   sequence.resize(N); // 不能少，为sequence创造N个空间pos.reserve(N); // 不能少//读入数据和初始化posfor(int i=0;i<N;i++) {scanf("%d",&sequence[i]);pos[sequence[i]]=i;}//temp储存circle中第一个位置的元素//swap_times记录swap(0,i)的次数//elements_num记录circle中元素的个数int temp，swap_times=0，elements_num=0;for(int element=0;element<N;element++){//temp0用来进行下面的赋值活动int element_circle=element; int temp0;temp=sequence[element_circle];//每次进入circle前初始化elements_num为0；elements_num=0; //若为单元环，则不进入while,故elements_num为0；while(pos[element_circle]!=element_circle){//判断当前位置sequence[i]是否为i，即是否是单元环和环是否结束；若不是，则继续if(sequence[element_circle]==0)        flag=1;//记录该circle是否有0 elements_num++; // 记录circle元素个数if(pos[pos[element_circle]]!=pos[element_circle])//用于判断当前元素是否环的尾巴sequence[element_circle]=element_circle; //不是尾巴elsesequence[element_circle]=temp;// 是尾巴，用temp来赋值//下面部分用来更新pos和移动element_circletemp0=element_circle;element_circle=pos[element_circle];pos[temp0]=temp0;//}if(elements_num!=0){ // 判断是否是单元环if(flag==1) swap_times+=elements_num-1; //判断circle是否有0，有0的非单元环移动elements-1次else swap_times+=elements_num+1; //无0的非单元环移动elements+1次}//归零flag和elements_numflag=elements_num=0;}cout<<swap_times<<end；return 0;
}