Pat1067:Sort with Swap(0,*)
1067. Sort with Swap(0,*) (25)
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 1Sample Output:
9
思路
贪心 + 桶排序。
1.用数组pos模拟N个桶记录输入的数的位置,数组的下标表示这个数。输入数字时根据输入次序i和数字大小num是否相等来决定需要和0交换的数字个数NumCount。
2.只要0没有在0位置,就将pos[0]与pos[pos[0]]交换.循环直到pos[0] == 0。
3.如果pos[0] == 0且还有数字不在正确的位置(NumCount > 0),那么就将0和最近的一个不在正确位置的数字交换。注意索引交换的数字时用index记录下标(不然有几个测试用例会超时),这样下一次pos[0] == 0需要索引新的交换数字时不用再重头开始。
4.用一个变量swaptimes统计交换次数,然后输出就行。
代码
#include<iostream> #include<vector> using namespace std; int main() { int N; while(cin >> N) { int index = 1,NumCount = 0,swaptimes = 0; vector<int> pos(N); for(int i = 0;i < N;i++) { int num; cin >> num; pos[num] = i; if(pos[num] != num && num != 0) NumCount++; } while(NumCount > 0) { if(pos[0] == 0) { while(index < N) { if(pos[index] != index) { swap(pos[0],pos[index]); swaptimes++; break; } index++; } } while(pos[0] != 0) { swap(pos[0],pos[pos[0]]); NumCount--; swaptimes++; } } cout << swaptimes << endl; } }