【题意+分析】1067 Sort with Swap(0, i) (25 分)

立志用最少的代码做最高效的表达

PAT甲级最优题解——>传送门

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

题意：每次只能以0所在的位置和某数交换，若想变为一个有序数列，问最小交换次数是多少。

算法逻辑：从贪心的角度看，每次都将0所在位置和一个不在正确位置的数字交换显然为最优解。那么，如果某次交换后，0回到了位置0上，就将其与随便一个不在正确位置上的数字交换即可。其中贪心思想请读者仔细体会。

#include<bits/stdc++.h>
using namespace std;
int pos[100010];
int main() {int n; cin >> n;for(int i = 0; i < n; i++) {int x; cin >> x;pos[x] = i;}int cnt = 0; for(int i = 0; i < n; i++) if(pos[i] != i) {while(pos[0] != 0) {swap(pos[0], pos[pos[0]]); cnt++;}if(pos[i] != i) {swap(pos[0], pos[i]);cnt++;}}cout << cnt; return 0;
}

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【题意+分析】1067 Sort with Swap(0, i) (25 分)_24行代码AC

立志用最少的代码做最高效的表达

耗时：

求赞哦~ (✪ω✪)

【题意+分析】1067 Sort with Swap(0, i) (25 分)_24行代码AC相关推荐

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