bzoj 2460: 元素 线性基
题目大意:
题解:
RT
线性基裸题
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
inline void read(int &x){
x=0;char ch;bool flag = false;
while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;
while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
inline void read(ull &x){
x=0;char ch;bool flag = false;
while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;
while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
const int maxn = 1024;
struct Node{
ull num,val;
bool friend operator < (const Node &a,const Node &b){
return a.val > b.val;
}
}a[maxn];
ull p[80];
inline bool check(ull x){
for(int i = 63;i>=0;--i){
if((x>>i)&1){
if(p[i] == 0){
p[i] = x;
return true;
}else{
x ^= p[i];
if(x == 0) return false;
}
}
}return false;
}
int main(){
int n;read(n);
for(int i=1;i<=n;++i){
read(a[i].num);
read(a[i].val);
}
sort(a+1,a+n+1);
int ans = 0;
for(int i=1;i<=n;++i){
if(check(a[i].num)) ans += a[i].val;
}printf("%d\n",ans);
getchar();getchar();
return 0;
}
人就像命运下的蝼蚁,谁也无法操控自己的人生.