该题目是EMAXX推荐的练习题,刘汝佳的书也有解说
如果S0属于全集,那S0就可以作为一个分组,那么S分组数可以是best{当前S中S0的补集+1}
对于集合类的题目我觉得有点抽象,希望多做多理解把
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define rrep(i,j,k) for(register int i=j;i>=k;i--)
#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])
#define iin(a) scanf("%d",&a)
#define lin(a) scanf("%lld",&a)
#define din(a) scanf("%lf",&a)
#define s0(a) scanf("%s",a)
#define s1(a) scanf("%s",a+1)
#define print(a) printf("%lld",(ll)a)
#define enter putchar(‘\n‘)
#define blank putchar(‘ ‘)
#define println(a) printf("%lld\n",(ll)a)
#define IOS ios::sync_with_stdio(0)
using namespace std;
const int maxn = 1e6+11;
const int oo = 0x3f3f3f3f;
const double eps = 1e-7;
typedef long long ll;
ll read(){
ll x=0,f=1;register char ch=getchar();
while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
return x*f;
}
int main(){
int kase=0,n,m,x,P[666],U[1<<17|1],dp[1<<17|1];
while(cin>>n){
if(n==0)break;
rep(i,0,n-1){
m=read();
P[i]=1<<i;
rep(j,1,m){
x=read();
P[i]|=(1<<x);
}
}
rep(S,0,(1<<n)-1){//并集枚举
U[S]=0;
rep(i,0,n-1){
if((S>>i)&1) U[S]|=P[i];
}
}
dp[0]=0; x=(1<<n)-1;//dp[S]:集合S最多可分成dp[S]组
rep(S,1,(1<<n)-1){
dp[S]=0;
for(int S0=S;S0;S0=(S0-1)&S){
if(U[S0]==x) dp[S]=max(dp[S],dp[S^S0]+1);
}
}
printf("Case %d: %d\n",++kase,dp[(1<<n)-1]);
}
return 0;
}