PAT (Advanced Level) Practise 1067 Sort with Swap(0,*) (25)
1067. Sort with Swap(0,*) (25)
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 1Sample Output:
9
暴力瞎跑，如果0不在0位置上，那把0和该放在这个位置的数交换，直到0回到0位置，然后看看有没有不在自己位置上的，
和0换了，在按照上面的步骤执行就ok了。
#include<cmath>
#include<queue>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 10;
int n, x, a[maxn], cnt, k = 0;
int main()
{
scanf("%d", &n);
cnt = 0;
for (int i = 0; i < n; i++)
{
scanf("%d", &x);
a[x] = i;
}
while (true)
{
while (a[0] != 0)
{
x = a[0];
cnt++;
swap(a[x], a[0]);
}
for (int i = k; i < n; i++)
if (a[i] != i)
{
swap(a[i], a[0]);
cnt++;
k = i;
break;
}
else if (i == n - 1)
{
printf("%d\n", cnt);
return 0;
}
}
return 0;
}