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Re:位运算,速度还将就,代码比较短In Reply To:位运算,速度还将就,代码比较短 Posted by:justryit at 2013-07-26 11:55:57 > #include<cstdio>
> #include<cstdlib>
> using namespace std;
> int ans=999999;
> int mi[16]={1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768};
> void search(int x,int f,int st)/*x表当前第几位,f表棋盘状态,st代表翻转次数*/
> {
> if((f==65535)or(f==0))
> {
> if(st<ans) ans=st;
> return;
> }
> if(x==17) return;
> if(st>=ans) return;
> search(x+1,f,st);
> f^=(1<<((17-x)-1));//改变自己
> if(x%4!=0) f^=(1<<((17-x-1)-1));//改变右边的
> if(x%4!=1) f^=(1<<((17-x+1)-1));//改变左边的
> if(x>4) f^=(1<<((17-x+4)-1));//改变上面的
> if(x<=12) f^=(1<<((17-x-4)-1));//改变下面的
> search(x+1,f,st+1);
> }
> int main()
> {
> int s=0;
> for(int i=1;i<=4;i++)
> {
> for(int j=1;j<=4;j++)
> {
> char ch;
> scanf("%c",&ch);
> if(ch=='b') s+=mi[16-4*i+4-j];/*把矩阵变成一个数,第i位为1代表第(i/4)行第(i%4)(0代表4)列是Black;*/
> }
> scanf("\n");
> }
> search(1,s,0);
> if (ans==999999) printf("Impossible\n");
> else printf("%d\n",ans);
> return 0;
> }
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