POJ - 1185 跑兵阵地状态压缩DP - c++编程基础

n()
{
int i,t,j,x,ans;
while (~scanf("%d%d",&n,&m))
{
anum=0;
a[0]=0;
for (i=0;i<(1< if (ok1(i))
{
a[++anum]=i;
cnt[anum]=0;
j=i;
while (j)
{
j=j&(j-1);
cnt[anum]++;
}
}
for (i=0;i<=n;i++) gets(arc[i]+1);
memset(dp,0,sizeof(dp));

for (i=1;i<=anum;i++)
if (ok2(1,a[i])) dp[1][0][i]=cnt[i];
for (t=2;t<=n;t++)
for (i=0;i<=anum;i++)
if (ok2(t,a[i]))
for (j=0;j<=anum;j++)
if (ok2(t-1,a[j]) && ok3(a[j],a[i]))
for (x=0;x<=anum;x++)
if (dp[t][j][i] dp[t][j][i]=dp[t-1][x][j]+cnt[i];

ans=0;
for (i=0;i<=anum;i++)
for (j=0;j<=anum;j++)
if (ans ans=dp[n][i][j];
printf("%d\n",ans);
}
return 0;
}

POJ - 1185 跑兵阵地 状态压缩DP (二)

POJ - 1185 跑兵阵地状态压缩DP (二)