使序列有序的最少交换次数(minimum swaps)
2017-02-22 07:12
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题目描述:(minimum swaps)
Given a sequence, we have to find the minimum no of swaps required to sort the sequence.
分析:
formula: no. of elements out of place - "cycles" in the sequence
A cycle is a set of elements, each of which is in the place of another. So in example sequences { 2, 1, 4, 3}, there are two cycles: {1, 2} and {3, 4}. 1 is in the place where 2 needs to go, and 2 is in the place where 1 needs to go 1, so they are a cycle;
likewise with 3 and 4.
The sequences {3, 2, 1, 5, 6, 8, 4, 7 }also has two cycles: 3 is in the place of 1 which is in the place of 3, so {1, 3} is a cycle; 2 is in its proper place; and 4 is in the place of 7 which is in the place of 8 in place of 6 in place of 5 in place of 4,
so {4, 7, 8, 6, 5} is a cycle. There are seven elements out of place in two cycles, so five swaps are needed.
实现:
e.g. { 2, 3, 1, 5, 6, 4}
231564 -> 6 mismatch
two cycles -> 123 and 456
swap 1,2 then 2,3, then 4,5 then 5,6 -> 4 swaps to sort
Probably the easiest algorithm would be to use a bitarray. Initialize it to 0, then start at the first 0. Swap the number there to the right place and put a 1 there. Continue until the current place holds the right number. Then move on to the next 0
Example:
231564
000000
-> swap 2,3
321564
010000
-> swap 3,1
123564
111000
-> continue at next 0; swap 5,6
123654
111010
-> swap 6,4
123456
111111
-> bitarray is all 1's, so we're done.
Given a sequence, we have to find the minimum no of swaps required to sort the sequence.
分析:
formula: no. of elements out of place - "cycles" in the sequence
A cycle is a set of elements, each of which is in the place of another. So in example sequences { 2, 1, 4, 3}, there are two cycles: {1, 2} and {3, 4}. 1 is in the place where 2 needs to go, and 2 is in the place where 1 needs to go 1, so they are a cycle;
likewise with 3 and 4.
The sequences {3, 2, 1, 5, 6, 8, 4, 7 }also has two cycles: 3 is in the place of 1 which is in the place of 3, so {1, 3} is a cycle; 2 is in its proper place; and 4 is in the place of 7 which is in the place of 8 in place of 6 in place of 5 in place of 4,
so {4, 7, 8, 6, 5} is a cycle. There are seven elements out of place in two cycles, so five swaps are needed.
实现:
e.g. { 2, 3, 1, 5, 6, 4}
231564 -> 6 mismatch
two cycles -> 123 and 456
swap 1,2 then 2,3, then 4,5 then 5,6 -> 4 swaps to sort
Probably the easiest algorithm would be to use a bitarray. Initialize it to 0, then start at the first 0. Swap the number there to the right place and put a 1 there. Continue until the current place holds the right number. Then move on to the next 0
Example:
231564
000000
-> swap 2,3
321564
010000
-> swap 3,1
123564
111000
-> continue at next 0; swap 5,6
123654
111010
-> swap 6,4
123456
111111
-> bitarray is all 1's, so we're done.
#include <iostream> #include <algorithm> using namespace std; template <typename T> int GetMinimumSwapsForSorted(T seq[], int n) { bool* right_place_flag = new bool ; T* sorted_seq = new T ; int p ,q; //// copy(seq, seq + n, sorted_seq); sort(sorted_seq, sorted_seq + n); ////可采用效率更高的排序算法 // 初始化bitarray for(int i = 0; i < n; i++) { if(seq[i] != sorted_seq[i]) right_place_flag[i] = false; else right_place_flag[i] = true; } //// p = 0; int minimumswap = 0; while(1) { while(right_place_flag[p]) p++; q = p + 1; // 在已排好序的数组中找到和未排好序的元素,此种找法只对无重复序列能得出minimum swaps while(q < n) { if(!right_place_flag[q] && sorted_seq[q] == seq[p]) break; q++; } if(q >= n || p >= n) break; right_place_flag[q] = true; // 对调后正好在排好序数组的位置上,则bitarray中的那位也置为true if(seq[q] == sorted_seq[p]) right_place_flag[p] = true; swap(seq[p], seq[q]); minimumswap++; } delete[] sorted_seq; delete[] right_place_flag; return minimumswap; } int _tmain(int argc, _TCHAR* argv[]) { int seq[] = {3, 2, 1, 5, 6, 8, 4, 7 };//{2,3,1,5,6,4};//{2,3,2,4,7,6,3,5}; int n = sizeof(seq) / sizeof(int); cout<<"minimum swaps : "<<GetMinimumSwapsForSorted(seq, n)<<endl; system("pause"); return 0; }
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