1067. Sort with Swap(0,*) (25)【贪心】——PAT (Advanced Level) Practise

题目信息

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

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 (<=10^5) 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

解题思路

首位为0，找出第一个未排序的数的位置进行交换

首位非0。与该数应处的位置交换

AC代码

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> a, b;
int main()
{
int n, t, c = 0;
scanf("%d", &n);
for (int i = 0; i < n; ++i){
scanf("%d", &t);
a.push_back(t);
}
int cnt = 0;
while (c < n){ //序列排序未结束
if (0 == a[0]){ //首位为0，找出第一个未排序的数的位置
for (; c < n && c == a[c]; ++c)
continue;
if (c >= n) break;
swap(a[0], a[c]);
++cnt;
}else{ //首位非0。与应处于的位置交换
swap(a[0], a[a[0]]);
++cnt;
}
}
printf("%d\n", cnt);
return 0;
}

个人游戏推广：



《10云方》与方块来次消除大战!