Source:

PAT A1067 Sort with Swap(0, i) (25 分)

Description:

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

Keys:

• 贪心

Code:

 1 /*
 2 Data: 2019-07-22 19:54:12
 3 Problem: PAT_A1067#Sort with Swap(0, i)
 4 AC: 29:14
 5 
 6 题目大意：
 7 排序，只能交换0和任意其他数，给出需要的最少步数
 8 
 9 基本思路：
10 1.0在几号位就跟相应的数字换位置，每次操作可以将一个数字归位；
11 2.若0回到了0号位，则再与任意一个不在自己位置上的数字交换；
12 3.重复第一步，直至所有数字都归位；
13 */
14 
15 #include<cstdio>
16 #include<algorithm>
17 using namespace std;
18 const int M=1e5+10;
19 int pos[M];
20 
21 int main()
22 {
23 #ifdef    ONLINE_JUDGE
24 #else
25     freopen("Test.txt", "r", stdin);
26 #endif
27 
28     int n,num,left=0,cnt=0,st=0;
29     scanf("%d", &n);
30     for(int i=0; i<n; i++)
31     {
32         scanf("%d", &num);
33         pos[num]=i;
34         if(num!=i && num!=0)
35             left++;
36     }
37     while(left!=0)
38     {
39         if(pos[0]==0){
40             for(int i=st; i<n; i++){
41                 if(pos[i]!=i){
42                     swap(pos[0],pos[i]);
43                     st=i;break;
44                 }
45             }
46             cnt++;
47         }
48         swap(pos[0],pos[pos[0]]);
49         cnt++;
50         left--;
51     }
52     printf("%d", cnt);
53 
54     return 0;
55 }