【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

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 typedef long long ll;
  4 typedef unsigned int uint;
  5 typedef unsigned long long ull;
  6 typedef long double ld;
  7 typedef pair<int,int> PII;
  8 typedef pair<ll,ll> Pll;
  9 typedef vector<int> VI;
 10 typedef vector<PII> VII;
 11 typedef pair<ll,ll>P;
 12 #define N  100010
 13 #define M  1000000
 14 #define INF 1e9
 15 #define fi first
 16 #define se second
 17 #define MP make_pair
 18 #define pb push_back
 19 #define pi acos(-1)
 20 #define mem(a,b) memset(a,b,sizeof(a))
 21 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 22 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 23 #define lowbit(x) x&(-x)
 24 #define Rand (rand()*(1<<16)+rand())
 25 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 26 #define ls p<<1
 27 #define rs p<<1|1
 28 #define fors(i) for(auto i:e[x]) if(i!=p)
 29 
 30 const int MOD=1e9+7,inv2=(MOD+1)/2;
 31       double eps=1e-6;
 32       int dx[4]={-1,1,0,0};
 33       int dy[4]={0,0,-1,1};
 34 
 35 int read()
 36 {
 37    int v=0,f=1;
 38    char c=getchar();
 39    while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
 40    while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
 41    return v*f;
 42 }
 43 
 44 ll readll()
 45 {
 46    ll v=0,f=1;
 47    char c=getchar();
 48    while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
 49    while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
 50    return v*f;
 51 }
 52 
 53 struct node
 54 {
 55     ll x;
 56     int y;
 57 }a[N];
 58 
 59 bool cmp(node a,node b)
 60 {
 61     return a.y>b.y;
 62 }
 63 
 64 struct base
 65 {
 66     ll d[65],p[65];
 67     int cnt;
 68 
 69     base()
 70     {
 71         mem(d,0);
 72         mem(p,0);
 73         cnt=0;
 74     }
 75 
 76     bool insert(ll x)
 77     {
 78         //if(x==0) return 0;
 79         per(i,62,0)
 80          if(x>>i&1)
 81          {
 82              if(!d[i])
 83              {
 84                  d[i]=x;
 85                  break;
 86              }
 87              x^=d[i];
 88          }
 89         return x>0;
 90     }
 91 }T;
 92 
 93 int main()
 94 {
 95     int n=read();
 96     rep(i,1,n) scanf("%lld%d",&a[i].x,&a[i].y);
 97     sort(a+1,a+n+1,cmp);
 98     base T;
 99     int ans=0;
100     rep(i,1,n) if(T.insert(a[i].x)) ans+=a[i].y;
101     printf("%d\n",ans);
102     return 0;
103 }

 

posted on 2019-11-05 15:26  myx12345  阅读(153)  评论(0)  编辑  收藏  举报

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