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：如果數(shù)字0當前在i號位上，則找到數(shù)字i當前所處的位置，然后把0與i進行交換則可以使有效交換增大 當0處于0號位置上時，使0與一個不在本位上的數(shù)交換，可以使無效交換次數(shù)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數(shù)的個數(shù) ，也即有效交換次數(shù) 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]);//當0元素在0位置時，使0元素與不在本位上的數(shù)交換，該交換為無效交換 ans++; break; } k++; } } while(pos[0]!=0){ swap(pos[0],pos[pos[0]]);//將pos[0]元素與0元素交換，使pos[0]元素回到其本位上，該交換是有效交換 ans++; //進行一次有效交換left減1 left--; } } printf("%d/n",ans); return 0;}新聞熱點
疑難解答
