PAT甲级1067

Posted 2022-08-30 wx630c98f24f6b8

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

时间限制

150 ms

内存限制

65536 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

/*
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
bool checksort(vector<int> a,int n,int &swapindex)

  for (int i = 0; i < n; i++)
  
    if (a[i] != i)
    
      swapindex = i;
      return false;
    
  
  return true;

int main()

  int N;
  scanf("%d", &N);
  //int *a = new int[N];
  //int *b = new int[N];
  int temp;
  int start;
  vector<int> a, b;
  for (int i = 0; i < N; i++)
  
    b.push_back(0);
  
  for (int i = 0; i < N; i++)
  
    scanf("%d", &temp);
    if (!temp)
      start = i;
    a.push_back(temp); b[temp] = i;
  
  int swapindex = 0,count=0;
  while (!checksort(a, N,swapindex))
  
    if (!start)
    
      int temp = a[swapindex];
      swap(a[swapindex], a[start]);
      b[0] = swapindex;
      b[temp] = start;
      count++;
      start = swapindex;
    
    while (start)
    
      for (int i = 0; i < N; i++)
      
        if (a[i] == start)
        
          swapindex = i;
          break;
        
      
      swap(a[swapindex], a[start]);
      count++;
      start = swapindex;
    
    while (start)
    
      swap(a[b[start]], a[start]);
      int temp = b[start];
      b[start] = start;
      b[0] = temp;
      start = temp;
      count++;
    
  
  printf("%d", count);
  return 0;

我的超时版本*/
#include <cstdio>
#include <vector>
using namespace std;
int main() 
  int n, num, cnt = 0, ans = 0, index = 1;
  scanf("%d", &n);
  vector<int> v(n);
  for (int i = 0; i < n; i++) 
    scanf("%d", &num);
    v[num] = i;//记录索引即可
    if (num != i && num != 0) cnt++;//记录索引与值不一致的个数
  
  while (cnt > 0) 
    if (v[0] == 0) 
      while (index < n) 
        if (v[index] != index) //记录上次处理时，第一次记录索引与值不一致时的索引
          swap(v[index], v[0]);//因为每次处理时都会使之一致，那么从这里开始往后
                    //进行遍历即可，不要从头开始遍历
          ans++;
          break;
        
        index++;
      
    
    while (v[0] != 0) 
      swap(v[v[0]], v[0]);
      ans++;
      cnt--;
    
  
  printf("%d", ans);
  return 0;

/*跟这位大神的思想一致，但对复杂度的处理不够好 ╮(╯▽╰)╭*/