c语言:金币阵列的问题

#include<stdio.h>

 const int inf = 99999;
 const int N = 101;

 int a[N][N],b[N][N],temp[N][N]; //a存储初始矩阵，b为目标状态矩阵
 int n,m;
 int need;//需要变换次数

 void ChangeL(int x,int y)//变换列
 {
     if(x==y)return;
     int i;
     for(i=1;i<=n;i++)
     {
         int tt=temp[i][y];
         temp[i][y]=temp[i][x];
         temp[i][x]=tt;
     }
     need++;
 }

 void ChangeH(int x)//变换行
 {
     int i;
     for(i=1;i<=m;i++)
     {
         temp[x][i]^=1;
     }
 }

 bool Same(int x,int y) //判断列是否满足条件
 {
     int i;
     for(i=1;i<=n;i++)
         if(b[i][x]!=temp[i][y])return false;
     return true;
 }

 int main()
 {
     int tests;
     scanf("%d",&tests); //数据组数

     while(tests--)
     {
         scanf("%d%d",&n,&m); //n行，m列
         int i,j;
         for(i=1;i<=n;i++)
             for(j=1;j<=m;j++)
             {
                 scanf("%d",&a[i][j]);
             }

             for(i=1;i<=n;i++)
                 for(j=1;j<=m;j++)
                     scanf("%d",&b[i][j]);

             int k;
             int ans=inf; //ans存储最终答案，初始值为无穷大

 
             for(k=1;k<=m;k++)//枚举各列为第一列
             {
                 for(i=1;i<=n;i++)
                     for(j=1;j<=m;j++)
                         temp[i][j]=a[i][j];
                 need=0;
                 ChangeL(1,k);

 
                 //不满足的行，进行一次变换
                 for(i=1;i<=n;i++)
                 {
                     if(temp[i][1]!=b[i][1])//该行不满足条件
                     {
                         ChangeH(i);//变换行
                         need++;
                     }
                 }

                 bool find;
                 for(i=1;i<=m;i++)//检查每列是否满足条件
                 {
                     find=false;
                     if(Same(i,i))
                     {
                         find=true;
                         continue;
                     }
                     for(j=i+1;j<=m;j++)//寻找temp中与b的i列相同的列
                     {
                         if(Same(i,j))//temp 的 j列于b的i列相同
                         {
                             if(Same(j,j))continue;//temp的j列与b的j列相同
                             ChangeL(i,j);//交换temp的i，j列
                             find=true;
                             break;
                         }
                     }
                     if(find==false)//找不到该列对应列
                     {
                         break;
                     }
                 }

                 if(find==true&&need<ans)
                     ans=need;
             }

             if(ans<inf)
                 printf("%d\n",ans);
             else
                 printf("-1\n");
     }
     return 0;
 }