1067 Sort with Swap(0, i) (25 分) 好，容易出错

2024-05-17 14:39:54

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:

注意：写题前一定要注意数组下标和对应内容的关系，然后循环体内的每一个变量的判定一定要仔细思考，这里一直卡在while（k<n）这里，没写一步循环将无限进行下去。使用swap函数时一定要加上头文件algorithm和using namespace std;

具体代码

#include <cstdio>
#include <algorithm>
using namespace std;
int main(){int n;scanf("%d",&n);int no=n-1;int pos[100010];int num;for(int i=0;i<n;i++){scanf("%d",&num);if(num==i&&num!=0)no--;pos[num]=i;//no用于存放除0外未回到原位的数字的数量}int k=1;int ans=0;while(no>0){//当还有未回归到原位的数字时if(pos[0]==0){while(k<n){if(pos[k]!=k){swap(pos[0],pos[k]);ans++;break;}k++;}}while(pos[0]!=0){swap(pos[0],pos[pos[0]]);ans++;no--;}
}
printf("%d\n",ans);
return 0;
}

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