Sorting a Three-Valued Sequence

Problem:

Sorting a Three-Valued Sequence 

Sorting is one of the most frequently performed computational tasks. Consider the special sorting problem in which the records to be sorted have at most threedifferent key values. This happens for instance when we sort medalists of a competition according to medal value, that is, gold medalists come first, followed by silver, and bronze medalists come last.

In this task the possible key values are the integers 1, 2 and 3. The required sorting order is non-decreasing. However, sorting has to be accomplished by a sequence of exchange operations. An exchange operation, defined by two position numbers p and q, exchanges the elements in positions p and q.

You are given a sequence of key values. Write a program that computes the minimal number of exchange operations that are necessary to make the sequence sorted.

PROGRAM NAME: sort3

INPUT FORMAT

Line 1:	N (1 <= N <= 1000), the number of records to be sorted
Lines 2-N+1:	A single integer from the set {1, 2, 3}

SAMPLE INPUT (file sort3.in)

9
2
2
1
3
3
3
2
3
1

OUTPUT FORMAT

A single line containing the number of exchanges required

SAMPLE OUTPUT (file sort3.out)

4

____________________________________________________________________________________________________________________________________

My Answer:

/*
ID:iby071
LANG:C++
TASK:sort3
*/

#include<iostream>
#include<fstream>
using namespace std;

int min(int a,int b)
{	if(a<b) return a;
	else return b;
}

int main()
{	ifstream fin("sort3.in");
	ofstream fout("sort3.out");
	int n,n1=0,n2=0,n3=0,error12=0,error21=0,error13=0,error31=0,error23=0,error32=0,high;
	int tmp,cnt=0;
	fin>>n;
	int *arr=new int[n];

	for(int i=0;i<n;++i)				//统计1,2,3的个数，已经相当于给出正确排序
	{	fin>>arr[i];
		switch(arr[i])
		{	case 1:++n1;break;
			case 2:++n2;break;
			case 3:++n3;break;
		}
	}

	for(int i=0;i<n1;++i)				//统计各类错误，其中errorij指应该放i的地方目前是j
	{	if(arr[i]==2) ++error12;
		if(arr[i]==3) ++error13;
	}
	high=n1+n2;
	for(int i=n1;i<high;++i)
	{	if(arr[i]==1) ++error21;
		if(arr[i]==3) ++error23;
	}
	high+=n3;
	for(int i=n1+n2;i<high;++i)
	{	if(arr[i]==1) ++error31;
		if(arr[i]==2) ++error32;
	}

				
	cnt+=min(error12,error21);			//统计可以通过一次交换归位的
	error12-=cnt;
	error21-=cnt;
	
	cnt+=min(error13,error31);
	error13-=cnt;
	error31-=cnt;
	
	cnt+=min(error23,error32);;
	error23-=cnt;
	error32-=cnt;

	cnt+=2*(error12+error21);			//不能一次交换归为的必然可以两次交换归位，剩余的error12=error23=error31,error21=error13=error32
	
	fout<<cnt<<endl;
	return 0;
}