HihoCoder - 1486

 https://cn.vjudge.net/problem/HihoCoder-1486

题意：给出每个物品的价值val和属性(1~m)（m<=10）;取一些使得这些的各种属性数都是奇数求最大价值；

思路：状态呀说dp;把属性压缩；异或后是各位是一就是奇数；要先初始化dp[i][0]=0;dp[i][j]=dp[i-1][j]每次多要更新的，后面会用到；

#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<iostream>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<sstream>
#include<cstdio>
#include<ctime>
#include<map>
#include<stack>
#include<string>
#include<ctime>
using namespace std;

#define sfi(i) scanf("%d",&i)
#define pri(i) printf("%d\n",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define mem(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define eps 1e-14
#define PI acos(-1)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))<eps)
#define fl() printf("flag\n")
#define MOD(x) ((x%mod)+mod)%mod
#define FASTIO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
ll gcd(ll a,ll b){while(b^=a^=b^=a%=b);return a;}
const int maxn=1e6+9;
const int mod=1e8+7;

inline ll read()
{
    ll f=1,x=0;
    char ss=getchar();
    while(ss<'0'||ss>'9'){if(ss=='-')f=-1;ss=getchar();}
    while(ss>='0'&&ss<='9'){x=x*10+ss-'0';ss=getchar();}
    return f*x;
}

int n,m;
int val[maxn],w[maxn];
int dp[1009][(1<<10)+10];
int main()
{
    //FASTIO;
    //#define endl '\n'

    int t;
    cin>>t;
    while(t--)
    {
        mem(val,0);
        mem(w,0);
        cin>>n>>m;
        for(int i=1;i<=n;i++)
        {
            int v,s;
            cin>>v>>s;
            val[i]=v;
            while(s--)
            {
                int k;
                cin>>k;
                w[i]=w[i]|(1<<(k-1));
            }
        }
        mem(dp,-1);
        //dp[1][w[1]]=val[1];
        for(int i=0;i<=n;i++)
        {
            dp[i][0]=0;
        }
        int ans=0;
        //cout<<dp[1][w[1]]<<endl;
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<(1<<m);j++)
            {
                dp[i][j]=dp[i-1][j];
                if(dp[i-1][j^w[i]]!=-1)
                    dp[i][j]=max(dp[i][j],dp[i-1][j^w[i]]+val[i]);
                ans=max(ans,dp[i][(1<<m)-1]);
                //cout<<i<<" "<<j<<" "<<dp[i][j]<<endl;
            }
        }
        cout<<ans<<endl;
    }
}