pat 1067

1067. Sort with Swap(0,*) (25)

时间限制

100 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

//两个测试点运行超时

#include <stdio.h>int a[100010];int n;void swap(int &a, int &b){  int tmp = b;  b = a;  a = tmp;}int main(){  int i, j, index, cnt, tag;  scanf("%d", &n);  for(i = 0; i < n; i++)  {    scanf("%d", &a[i]);    if(a[i] == 0)      index = i;  }  cnt = 0;  while(true)  {    if(a[0] != 0)    {      for(i = 0; i < n; i++)      {        if(a[i] == index)        {          swap(a[i], a[index]);          index = i;          cnt++;          i = 0;        }      }    }    if(a[0] == 0)    {      for(i = 1; i < n; i++)      {        if(a[i] != i)        {          swap(a[i], a[0]);          index = i;          cnt++;          break;        }      }          if(i == n) break;    }  }  printf("%d\n", cnt);  return 0;}