[BJWC2011]元素 线性基

Posted 2021-02-08 ww3113306

题面

题面

题解

一个方案合法，当且仅当选取的01串凑不出0.
因此就是要使得选取的01串全在线性基内，具体原因可以看这道题：[CQOI2013]新Nim游戏 线性基
要使得魔力值最大，只需要按法力值从大到小，贪心的往线性基中加串就可以了

#include<bits/stdc++.h>
using namespace std;
#define R register int
#define AC 1100
#define LL long long

int n; LL ans;
LL f[AC];
struct node{LL x, w;}s[AC];
inline bool cmp(node a, node b){return a.w > b.w;}

inline LL read()
{
    LL x = 0;char c = getchar();
    while(c > '9' || c < '0') c = getchar();
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x;
}

void pre()
{
    n = read();
    for(R i = 1; i <= n; i ++) s[i].x = read(), s[i].w = read();
    sort(s + 1, s + n + 1, cmp);
}

void work()
{
    for(R i = 1; i <= n; i ++)//从高开始贪心
    {
        LL x = s[i].x, maxn = 1LL << 60; bool done = false;
        for(R j = 60; ~j; j --, maxn >>= 1)
        {
            if(!(x & maxn)) continue;
            if(!f[j]) {f[j] = x, done = true; break;}
            else x ^= f[j];
        }
        if(done) ans += s[i].w;//加入线性基就要拿走
    }
    printf("%lld
", ans);
}

int main()
{
//  freopen("in.in", "r", stdin);
    pre();
    work();
//  fclose(stdin);
    return 0;
}