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再来一个25行的插头DP的AC代码，具体就不解释了XD

Posted by Sayakiss at 2010-08-10 09:57:09 on Problem 2411
In Reply To:25行代码流出 Posted by:Sayakiss at 2010-08-09 18:51:37

#include<iostream>
#include<cstring>
using namespace std;
long long f[2][1<<12],i,j,ps,p=0,c=1,n,m;
int main()
{
	while (scanf("%d%d",&n,&m),n!=0)
	{
	memset(f[c],0,sizeof(f[c]));
	f[c][(1<<m)-1]=1;
	for (i=0;i<n;i++)
		for (j=0;j<m;j++)
		{
			swap(p,c);memset(f[c],0,sizeof(f[c]));
			for (ps=0;ps<1<<m;ps++)
			if (f[p][ps])
			{
				if (j>0&&(!(ps&(1<<j-1)))&&(ps&(1<<j))) f[c][ps|1<<(j-1)]+=f[p][ps];
				if (!(ps&1<<j)) f[c][ps|1<<j]+=f[p][ps];
				if (ps&1<<j) f[c][ps^1<<j]+=f[p][ps];
			}
		}
	printf("%lld\n",f[c][(1<<m)-1]);
	}
}

Followed by:

Re:再来一个25行的插头DP的AC代码，具体就不解释了XD (12) tangbohu222 2011-03-23 18:45:18
看懂了一点 (86) cotton 2011-07-21 20:57:37
Re:再来一个25行的插头DP的AC代码，具体就不解释了XD (616) www123 2011-07-29 20:10:53
Re:再来一个25行的插头DP的AC代码，具体就不解释了XD (634) Naive_nico 2011-08-15 21:30:08

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