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 (≤) 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

题目分析：看浙大《数据结构》的时候见到过这题 重做并没有做成功 运行超时了 我想的是每次通过0与0的位置来归位 若0在这个过程中不小心被交换到了0的位置 那就得将0交换到还未被归为的那个元素上 思路是对的 做法导致时间复杂度过大
通过上面的分析 可以将0所在的看成一个环 只需要记录环中有多少元素 以及有多少环 当0所在的环计算完成后 就到另一个环去

过了3个测试点

 #define _CRT_SECURE_NO_WARNINGS
 #include <climits>
 #include<iostream>
 #include<vector>
 #include<queue>
 #include<map>
 #include<set>
 #include<stack>
 #include<algorithm>
 #include<string>
 #include<cmath>
 using namespace std;
 int Address[];
 int Array[];
 void swap(int i, int j)//交换2个地址上的值 i，j为地址
 {
     Address[Array[i]] = j;
     Address[Array[j]] = i;
     int temp = Array[i];
     Array[i] = Array[j];
     Array[j] = temp;
 }
 int main()
 {
     int N;
     int times = ;
     cin >> N;
     for (int i = ; i < N; i++)
     {
         cin >> Array[i];
         Address[Array[i]] = i;
     }
     int flag = ;
     while (flag)
     {
         if (Address[] == ){
             for (int i = ; i < N; i++)
                 if (Address[i] != i) {
                     swap(Address[], Address[i]);
                     flag = ;
                     times++;
                     break;
                 }
                 else flag = ;
         }
         else{
             swap(Address[], Address[Address[]]);
             times++;
         }
     }
     cout << times;
 }

全过

 #define _CRT_SECURE_NO_WARNINGS
 #include<stdio.h>

 int A[] = {  };
 int Position[] = {  };
 int IsRight[] = {  };
 void Swap(int i, int j)
 {
     int tmp = A[i];
     A[i] = A[j];
     A[j] = tmp;
 }
 int SwapTimes=;
 int FindElements(int Pos)
 {
     int num=;
     while (Position[Pos]!=Pos)
     {
         num++;
         IsRight[Position[Pos]] = ;
         Position[Pos] = Position[Position[Pos]];
     }
     return num;  //返回元素的个数
 }
 void Charge(int N)
 {
     int num;
     //从零开始计算
     SwapTimes += FindElements()-;            //交换次数比元素个数少一
     for (int i = ; i < N; i++)
     {
         if (!IsRight[i])
             SwapTimes += FindElements(i)+;        //虽然交换次数比元素个数少一 但是要利用0来进行交换 所以时 这个环的元素加一
     }                                            //而把0元素添加到 环中先进行一次交换 所以 最后结果为  元素+1-1+1
 }
 int main()
 {
     int N;
     scanf("%d", &N);
     for (int i = ; i < N; i++)
     {
         int num;
         scanf("%d", &num);
         A[i] = num;
         Position[num] = i;
         if (A[i] == i)
             IsRight[i] = ;
     }
     Charge(N);
     printf("%d", SwapTimes);
     return ;
 }