首先,先贴柳神的博客
想要刷好PTA,强烈推荐柳神的博客,和算法笔记
题目原文
1067 Sort with Swap(0, i) (25分)
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
生词如下:
nonnegative 非负
题目大意
有一个从0到n-1的数组被打乱了,现在要你重现排列,只能用0来两两交换.现在要你给出最少的交换次数.
题目思路
① 0只要不在开始的位置上,就把0和他位置上的人交换,
② 0一旦回到了开始的位置上,就把0和最近的一个(不是正确的数)交换.
柳神的代码
#include <iostream>
using namespace std;
int main() {
int n, t, cnt = 0, a[100010];
cin >> n;
for (int i = 0; i < n; i++) {
cin >> t;
a[t] = i;
}
for (int i = 1; i < n; i++) {
if (i != a[i]) {
while (a[0] != 0) {
swap(a[0], a[a[0]]);
cnt++;
}
if (i != a[i]) {
swap(a[0], a[i]);
cnt++;
}
}
}
cout << cnt;
return 0;
}
反思
① 我又犯了老毛病,复杂的问题简单化,我一开始想的是,交换的次数是一定的,然后就想着不交换偷懒.后来自己手动模拟了才发现,根本不是这样的.
② 交换的此时,以及0回到原文的次数都是不固定的,哪里有那么简单的题目嘛