10-排序6

10-排序6 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

题目没看仔细，是0-N-1，不用排序的，所以给排了序。自己开始写的代码如下：

#include<iostream>
#include<algorithm>
using namespace std;
int n,last = 0;
struct book{
    int sorted;
    int index;
};
int cmp(book a,book b){
    return a.sorted < b.sorted;
}

int findNext(book books[]){
    while(last < n && books[last].sorted < 0) last++;
    return last;
}
int main(){
    int i,next,cnt = 0,cntNoMove = 0;
    cin>>n;
    if(n == 1){
        cout<<0<<endl;
        return 0;
    }
    book books[n];
    int A[n];
    for(i=0;i<n;i++){
        cin>>A[i];
        books[i].sorted = A[i];
        books[i].index = i;
    }
    sort(books,books + n,cmp);
    for(i=0;i<n;i=next){
        if(books[i].sorted == A[i]){
            books[i].sorted = -2;
            next = findNext(books);
            cntNoMove++;
        }
        else if(books[i].sorted >= 0){
            books[i].sorted = -1;
            next = books[i].index;
        }
        else if(books[i].sorted == -1){
            cnt++;
            next = findNext(books);
            last = next;
            if(next == n) break;
        }
    }
    if(A[0] == 0) cout<<n - cntNoMove + cnt<<endl;
    else cout<<n - cntNoMove + cnt - 2<<endl;
    return 0;
}

借鉴别人的代码如下：

#include<stdio.h>
#define Max 100000

int main()
{
    int A[Max],Table[Max],flag[Max],N;
    int i,tmp,S,K;
    S=K=0;
    scanf("%d",&N);
    for (i=0;i<N;i++)
    {
        scanf("%d",&A[i]);
        flag[i]=0;   //标识元素访问过了没有
    }
    /* 指针数组，用来存放正确的序号 */
    for (i=0;i<N;i++)
    {
        Table[A[i]]=i;  //即元素A[i]存放在序号i中
    }

    for (i=0;i<N;i++)
    {
        if (flag[i]==0)
        {
            if (Table[i]!=i)
            {
                flag[i]=1;
                tmp=Table[i];
               while(flag[tmp]==0)
               {
                    flag[tmp]=1;
                    tmp=Table[tmp];
               }
               K++;
            }
            else if (Table[i]==i)
            {
                flag[i]=1;
                S++;
            }
        }
    }
    if (A[0]==0)printf("%d",N-S+K);
    else if (S==N)printf("0");
    else printf("%d",N-S+K-2);
    return 0;
}