10-排序6 Sort with Swap(0, i) (25 分)
Posted wanghao-boke
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了10-排序6 Sort with Swap(0, i) (25 分)相关的知识,希望对你有一定的参考价值。
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
#include<cstdio> #include<algorithm> using namespace std; const int maxn = 100010; int pos[maxn]; int main(){ int n,ans = 0; scanf("%d",&n); int left = n - 1,num; for(int i = 0; i < n; i++){ scanf("%d",&num); pos[num] = i; if(num == i && num != 0){ left--; } } int k = 1; while(left > 0){ if(pos[0] == 0){ while(k < n){ if(pos[k] != k){ swap(pos[0],pos[k]); ans++; break; } k++; } } if(pos[0] != 0){ swap(pos[0],pos[pos[0]]); ans++; left--; } } printf("%d",ans); return 0; }
以上是关于10-排序6 Sort with Swap(0, i) (25 分)的主要内容,如果未能解决你的问题,请参考以下文章
PAT甲级--Sort with Swap(0,*) (25)--下标数组交换法
1067 Sort with Swap(0, i) (25 分)