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 (≤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
心得:题意是给出一段序列,用每次让0和其他数字交换位置的方法使序列变成非第减序列,求交换的次数。
哨兵的使用:本题中0作为排序交换的哨兵,通过交换实现序列有序。
实现方法 :1、用数组记录每个数所在的位置,然后交换a[0],和a[a[0]],让0原来在的位置上的数字归位。
直到0回到0的位置上。
2、交换后的序列有两种请况,一是每个位置对应相应的点,即排好序了;二是还有的位置没有交换,则以0位哨兵重复1的操作。
#include<iostream>
using namespace std;
int a[100100];
void sp(int x,int y)
{
int tp=a[x];
a[x]=a[y];
a[y]=tp;
}
int main(void)
{
int n,i,t;
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%d",&t);
a[t]=i;
}
int cnt=0;
for(i=0;i<n;i++)
{
while(a[0]!=0)
{
sp(a[0],a[a[0]]);
cnt++;
}
if(a[i]!=i)
{
sp(a[0],a[i]);
cnt++;
}
}
printf("%d\n",cnt);
return 0;
}