UVA11825 Hackers ' crackdown

Exercises

The first pressure DP problem. So write it carefully.

The title translates to:

There are n sets grouped to find the number of groups that satisfy the condition

The status is: N sets, with each collection selected or not selected. A total of 1<<n species

According to the binary, can be pushed forward backwards F[s]=max (F[s-s0]) +1, S0 is a subset of S, Cover[s0] equals the complete

Subset enumeration:

 for (int x = S; x; x = (x1) &s)

Explain

Http://www.cnblogs.com/jffifa/archive/2012/01/16/2323999.html

(x-1) &s 0 of S is ignored, and the result is continuously reduced by 1

Code:

#include <bits/stdc++.h>using namespacestd;Const intmaxn=1<< -;intN,m,x,all,case=1;intcover[maxn],f[maxn],p[ -];intMain () { while(~SCANF ("%d", &n) &&N) { for(intI=0; i<n;i++) {P[i]=1<<i; scanf ("%d",&m);  while(m--) {scanf ("%d",&x); P[i]|=(1&LT;&LT;X);//The collection of each point and the adjacent point is represented by a binary, with n points, so there are n sets}} All=(1<<n)-1;  for(ints=0;s< (1&LT;&LT;N); s++) {Cover[s]=0;  for(intI=0; i<n;i++)            {                if(s& (1<<i)) Cover[s]|=p[i];//s represents the selected set of binary and, cover[s] to indicate how many points are selected for these collections, and the binary representation            }        }         for(ints=0;s< (1&LT;&LT;N); s++) {F[s]=0;  for(intS0=s;s0;s0= (s0-1) &s)//S0 as a subset of S            {                if(Cover[s0]==all) F[s]=max (f[s],f[s^s0]+1); }} printf ("Case %d:%d\n", case++, F[all]); }    return 0;}

UVA11825 Hackers ' crackdown