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 (<=105) 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 算法思想：如果数字0当前在i号位上，则找到数字i当前所处的位置，然后把0与i进行交换则可以使有效交换增大 当0处于0号位置上时，使0与一个不在本位上的数交换，可以使无效交换次数最小
#include<cstdio>#include<algorithm>using namespace std;const int maxn=100010;int pos[maxn];int main(){ int n; scanf("%d",&n); int left=n-1,num;//left存放除零以外不在本位上的数的个数 ，也即有效交换次数 for(int i=0;i<n;i++){ scanf("%d",&num); pos[num]=i; if(num==pos[num]&&num!=0) left--; } int ans=0; int k=1; while(left>0){ if(pos[0]==0){ while(k<n){ if(pos[k]!=k){ swap(pos[0],pos[k]);//当0元素在0位置时，使0元素与不在本位上的数交换，该交换为无效交换 ans++; break; } k++; } } while(pos[0]!=0){ swap(pos[0],pos[pos[0]]);//将pos[0]元素与0元素交换，使pos[0]元素回到其本位上，该交换是有效交换 ans++; //进行一次有效交换left减1 left--; } } printf("%d/n",ans); return 0;}新闻热点
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