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状压DP,有不太详细的注释,719ms#include<cstdio> #include<cstring> const long long N=12; long long n,m; long long f[N][(1<<12)];//第i行的状态为j,i-1行都已经全部填满有多少种状态 long long tt; bool ooo[(1<<12)]; //0:没有覆盖 1:已经覆盖 bool check (long long x,long long y)//x和y是否可以相得益彰 { if ((x|y)!=tt-1) return false; long long kk=(x&y); return ooo[kk]; } void prepare () { memset(ooo,true,sizeof(ooo)); for (long long u=0;u<(1<<12);u++) { long long x=u; bool last=false; while (x>0) { if ((x&1)==1) last=!last; else if (last==true) { ooo[u]=false; break; } x=x>>1; } if (last==true) ooo[u]=false; } return ; } int main() { prepare(); while (true) { memset(f,0,sizeof(f)); scanf("%I64d%I64d",&n,&m); if (n==0&&m==0) break; if (((n*m)&1)==1) {printf("0\n");continue;} tt=(1<<m); for (long long u=0;u<tt;u++) if (ooo[u]==true) f[1][u]=1; for (long long u=2;u<=n;u++) for (long long j=0;j<tt;j++)//枚举状态 for (long long k=0;k<tt;k++)//枚举上一行的状态 if (check(j,k)==true) f[u][j]+=f[u-1][k]; printf("%I64d\n",f[n][(tt-1)]); } return 0; } Followed by: Post your reply here: |
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