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

 1 #define _CRT_SECURE_NO_WARNINGS
 2 #include <climits>
 3 #include<iostream>
 4 #include<vector>
 5 #include<queue>
 6 #include<map>
 7 #include<set>
 8 #include<stack>
 9 #include<algorithm>
10 #include<string>
11 #include<cmath>
12 using namespace std;
13 int Address[100001];
14 int Array[100001];
15 void swap(int i, int j)//交换2个地址上的值 i，j为地址
16 {
17     Address[Array[i]] = j;
18     Address[Array[j]] = i;
19     int temp = Array[i];
20     Array[i] = Array[j];
21     Array[j] = temp;
22 }
23 int main()
24 {
25     int N;
26     int times = 0;
27     cin >> N;
28     for (int i = 0; i < N; i++)
29     {
30         cin >> Array[i];
31         Address[Array[i]] = i;
32     }
33     int flag = 1;
34     while (flag)
35     {
36         if (Address[0] == 0){
37             for (int i = 1; i < N; i++)
38                 if (Address[i] != i) {
39                     swap(Address[0], Address[i]);
40                     flag = 1;
41                     times++;
42                     break;
43                 }
44                 else flag = 0;
45         }
46         else{
47             swap(Address[0], Address[Address[0]]);
48             times++;
49         }            
50     }
51     cout << times;
52 }

View Code