立志用最少的代碼做最高效的表達

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PAT甲級最優(yōu)題解——>傳送門

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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

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題意：每次只能以0所在的位置和某數(shù)交換，若想變?yōu)橐粋€有序數(shù)列，問最小交換次數(shù)是多少。

算法邏輯：從貪心的角度看， 每次都將0所在位置和一個不在正確位置的數(shù)字交換顯然為最優(yōu)解。 那么，如果某次交換后，0回到了位置0上，就將其與隨便一個不在正確位置上的數(shù)字交換即可。 其中貪心思想請讀者仔細(xì)體會。

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#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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耗時：

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求贊哦~ (?ω?)

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