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1 3 1 2 3
2 Hint: If we choose 2 and 3,one is not divisible by the other,which is the most number you can choose.
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
typedef pair<int,int>pil;
const int maxn = 1000+5;
const int maxnnode=maxn*maxn;
const int mod = 1000000007;
struct DLX{
int n,m,size;
int U[maxnnode],D[maxnnode],L[maxnnode],R[maxnnode],Row[maxnnode],Col[maxnnode];
int H[maxn],S[maxn];//H[i]位置,S[i]个数
int ansd;
void init(int a,int b)
{
n=a; m=b;
REPF(i,0,m)
{
S[i]=0;
U[i]=D[i]=i;
L[i]=i-1;
R[i]=i+1;
}
R[m]=0; L[0]=m;
size=m;
REPF(i,1,n)
H[i]=-1;
}
void link(int r,int c)
{
++S[Col[++size]=c];
Row[size]=r;
D[size]=D[c];
U[D[c]]=size;
U[size]=c;
D[c]=size;
if(H[r]<0) H[r]=L[size]=R[size]=size;
else
{
R[size]=R[H[r]];
L[R[H[r]]]=size;
L[size]=H[r];
R[H[r]]=size;
}
}
void remove(int c)
{
for(int i=D[c];i!=c;i=D[i])
L[R[i]]=L[i],R[L[i]]=R[i];
}
void resume(int c)
{
for(int i=U[c];i!=c;i=U[i])
L[R[i]]=R[L[i]]=i;
}
bool v[maxn];
int f()
{
int ret = 0;
for(int c = R[0];c != 0;c = R[c])v[c] = true;
for(int c = R[0];c != 0;c = R[c])
if(v[c])
{
ret++;
v[c] = false;
for(int i = D[c];i != c;i = D[i])
for(int j = R[i];j != i;j = R[j])
v[Col[j]] = false;
}
return ret;
}
void Dance(int d)
{
// if(d + f() >=ansd) return ;
if(R[0] == 0)
{
if(d>ansd) ansd=d;
return ;
}
int c = R[0];
for(int i = R[0];i != 0;i = R[i])
if(S[i] < S[c])
c = i;
for(int i = D[c];i != c;i = D[i])
{
remove(i);
for(int j = R[i];j != i;j = R[j])remove(j);
Dance(d+1);
for(int j = L[i];j != i;j = L[j])resume(j);
resume(i);
}
}
};
DLX L;
LL num[maxn];
int t,n;
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
L.init(n,n);
REPF(i,1,n) scanf("%I64d",&num[i]);
REPF(i,1,n)
{
REPF(j,1,n)
if(num[i]%num[j]==0||num[j]%num[i]==0) L.link(i,j);
}
L.ansd=0;
L.Dance(0);
printf("%d\n",L.ansd);
}
return 0;
}
HDU 3335 Divisibility(DLX可重复覆盖)
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原文地址:http://blog.csdn.net/u013582254/article/details/42155203