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位运算,速度还将就,代码比较短#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; } Followed by:
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