PAT (Advanced Level) Practise 1067 Sort with Swap(0,*) (25)

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

题意：给你n个数，分别为0~n-1，每次可以将0和其他数字进行交换，问最少要交换几次，使得n个数字最后的顺序为0~n-1

解题思路：若0不在0号位，则将应该在0所在位置的数移过来，直至0到0号位，然后判断所有数所在的位置是否正确，若有不正确的，则用0和它交换

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int a[100009], n,x;

int main()
{
while (~scanf("%d", &n))
{
for (int i = 0; i < n; i++) scanf("%d", &x),a[x]=i;
int ans = 0,k=0;
while (1)
{
while (a[0] != 0)
{
ans++;
x = a[0];
swap(a[x], a[0]);
}
int flag = 1;
for (int i = k; i < n; i++)
{
if (i != a[i])
{
flag = 0; ans++;
swap(a[i], a[0]);
k = i;break;
}
}
if (flag) { printf("%d\n", ans); break; }
}
}
return 0;
}