p4570 [BJWC2011]元素

Posted 2021-01-27 yzxverygood

传送门

分析

对法力值从大到小排序然后对编号跑线性基即可

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<ctime>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
struct node {
    long long sum;
    int val;
};
node d[1100];
long long belong[110];
inline bool cmp(node x,node y){return x.val>y.val;}
int main(){
    int n,m,i,j,Ans=0;
    scanf("%d",&n);
    for(i=1;i<=n;i++)scanf("%lld%d",&d[i].sum,&d[i].val);
    sort(d+1,d+n+1,cmp);
    for(i=1;i<=n;i++){
      long long k=d[i].sum;
      for(j=60;j>=0;j--)
        if((1ll<<j)&k){
          if(!belong[j]){
              belong[j]=k;
            Ans+=d[i].val;
              break;
          }
          k^=belong[j];
        }
    }
    printf("%d
",Ans);
    return 0;
}