【BZOJ2460】元素（拟阵）

题意：给定n个物品，每个物品有属性x和价值y，要求从中选出一些使得价值和最大并且其中没有属性xor和为0的非空子集

n<=1000，x<=1e18，y<=1e4

思路：没有xor和为0的非空子集本来就是线性基的定义

拟阵，直接按价值排序之后贪心插入并维护线性基

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef long double ld;
 typedef pair<int,int> PII;
 typedef pair<ll,ll> Pll;
 typedef vector<int> VI;
 typedef vector<PII> VII;
 typedef pair<ll,ll>P;
 #define N  100010
 #define M  1000000
 #define INF 1e9
 #define fi first
 #define se second
 #define MP make_pair
 #define pb push_back
 #define pi acos(-1)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 #define lowbit(x) x&(-x)
 #define Rand (rand()*(1<<16)+rand())
 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 #define ls p<<1
 #define rs p<<1|1
 #define fors(i) for(auto i:e[x]) if(i!=p)

 const int MOD=1e9+,inv2=(MOD+)/;
       double eps=1e-;
       int dx[]={-,,,};
       int dy[]={,,-,};

 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 ll readll()
 {
    ll v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 struct node
 {
     ll x;
     int y;
 }a[N];

 bool cmp(node a,node b)
 {
     return a.y>b.y;
 }

 struct base
 {
     ll d[],p[];
     int cnt;

     base()
     {
         mem(d,);
         mem(p,);
         cnt=;
     }

     bool insert(ll x)
     {
         //if(x==0) return 0;
         per(i,,)
          if(x>>i&)
          {
              if(!d[i])
              {
                  d[i]=x;
                  break;
              }
              x^=d[i];
          }
         return x>;
     }
 }T;

 int main()
 {
     int n=read();
     rep(i,,n) scanf("%lld%d",&a[i].x,&a[i].y);
     sort(a+,a+n+,cmp);
     base T;
     int ans=;
     rep(i,,n) if(T.insert(a[i].x)) ans+=a[i].y;
     printf("%d\n",ans);
     return ;
 }