UVA11825Hackers' Crackdown

Posted 2020-08-21

题解:

首先将相邻点进行二进制，保存在p[i]中

然后将不同组合的p[i]组合的值记录下来，保存在cover[i]中

然后从小到大进行dp

s0为集合s的子集

if(cover[ s0 ] == all - 1) f[s] = max( f[s], f[s ^ s0] + 1);

代码:

#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define se second
#define fs first
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define pii pair<int,int>
const int INF = 1000000000;
const int maxn = 1 << 16;

int n, m, x;
int p[ 16 ], cover[ maxn ];
int f[ maxn ];

int main()
{
    int Case = 1;
    while( ~scanf( "%d", &n ) && n )
    {
        memset( p, 0, sizeof( p ) );
        for( int i = 0; i < n; i ++ )
        {
           scanf( "%d", &m );
           p[ i ] = ( 1 << i );
           while( m -- )
           {
               scanf( "%d", &x );
               p[ i ] |= ( 1 << x );
           }
        }
        int ALL = 1 << n;
        for( int s = 0; s < ALL; s ++ )
        {
            cover[ s ] = 0;
            for( int i = 0; i < n; i ++ )
            if( s & ( 1 << i ) ) cover[ s ] |= p[ i ];
        }
        for( int s = 0; s < ALL; s ++ )
        {
            f[ s ] = 0;
            for( int s0 = s; s0; s0 = ( s0 - 1 ) & s )
            {
                if( cover[ s0 ] == ALL - 1 ) f[ s ] = max( f[ s ], f[ s ^ s0 ] + 1 );
            }
        }
        printf( "Case %d: %d\n", Case ++, f[ ALL - 1 ] );
    }
    return 0;
}