USACO chapter 2 section 2.1 Sorting A Three-Valued Sequence

USACO chapter 2 section 2.1 Sorting A Three-Valued Sequence

USER: tianbing tianbing [tbbd4261]
TASK: sort3
LANG: C++

Compiling...
Compile: OK

Executing...
   Test 1: TEST OK [0.011 secs, 2932 KB]
   Test 2: TEST OK [0.000 secs, 2932 KB]
   Test 3: TEST OK [0.022 secs, 2932 KB]
   Test 4: TEST OK [0.000 secs, 2932 KB]
   Test 5: TEST OK [0.011 secs, 2932 KB]
   Test 6: TEST OK [0.022 secs, 2932 KB]
   Test 7: TEST OK [0.000 secs, 2932 KB]
   Test 8: TEST OK [0.000 secs, 2932 KB]

All tests OK.

Your program ('sort3') produced all correct answers!  This is your
submission #5 for this problem.  Congratulations!

用cnt[i]纪录i的数目，数目确定了i应该在的范围也就确定了
用s[i][j]表示应该出现i的范围内中含有j的数目
1  如果i中有j，j中有i，直接交换即可，交换一次
2  第一种情况交换完以后，如果 1中有2 ，2中有3，则3中必然有1 这种情况需交换2次才能排好序
                                     或者反过来，2中有1，3中有2, 1中有3，同样交换两次 

/*
ID:tbbd4261
PROG:sort3
LANG:C++
*/

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

ifstream fin("sort3.in");
ofstream fout("sort3.out");

int arr[1005]={0};
int cnt[4]={0};
int s[4][4]={0};

int min(int a,int b)
{
    return a>b?b:a;
}

int main()
{
   int n,i,j;
   fin>>n;
   for(i=1; i<=n; i++)
   {
      fin>>arr[i];
      cnt[arr[i]]++;
   }
  
   for(i=1; i<=cnt[1]; i++)
       s[1][arr[i]]++;
   for(i=cnt[1]+1; i<=cnt[1]+cnt[2]; i++)
       s[2][arr[i]]++;
   for(i=cnt[1]+cnt[2]+1; i<=n; i++)
        s[3][arr[i]]++;
   
   int ans=0;
   for(i=1; i<=3; i++)
   for(j=1; j<=3; j++)
   {
            if(i==j)continue;
            int temp=min(s[i][j],s[j][i]);
            ans+=temp; 
            s[i][j]-=temp; s[j][i]-=temp;
            
   }
   
   ans+=min(s[1][2],min(s[2][3],s[3][1]))*2;
   ans+=min(s[3][2],min(s[2][1],s[1][3]))*2;
   fout<<ans<<endl;
      
   //system("pause");
    return 0;
}