标签:contest define print can turn space str get scanf
题意
给定序列,$q(1\leq q \leq 100000) $次询问,每次查询给定区间内的最大异或子集。
思路
涉及到最大异或子集肯定从线性基角度入手。将询问按右端点排序后离线处理询问,对线性基的每一位贪心的保留靠后的。
代码
#include <bits/stdc++.h>
#define DBG(x) cerr << #x << " = " << x << endl;
using namespace std;
const int N = 500000 + 5;
int n, m, a[N], ans[N], L = 1;
struct node {
int l, r, id;
bool operator < (const node & a) const {
if(r != a.r) return r < a.r;
return l < a.l;
}
}b[N];
struct base {
int b[21], pos[21];
void add(int x, int id) {
for(int i = 20; i >= 0; i--) {
if((x >> i) & 1) {
if(!pos[i]) {
b[i] = x, pos[i] = id;
return;
}
if(id > pos[i]) {swap(x, b[i]); swap(id, pos[i]);}
x ^= b[i];
}
}
}
int query(int id) {
int res = 0;
for(int i = 20; i >= 0; i--) {
if(pos[i] >= id && (res ^ b[i]) > res) res ^= b[i];
}
return res;
}
}bs;
int main() {
scanf("%d", &n);
for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
scanf("%d", &m);
for(int i = 1; i <= m; i++) scanf("%d%d", &b[i].l, &b[i].r), b[i].id = i;
sort(b + 1, b + 1 + m);
for(int i = 1; i <= m; i++) {
while(L <= b[i].r) bs.add(a[L], L), L++;
ans[b[i].id] = bs.query(b[i].l);
}
for(int i = 1; i <= m; i++) printf("%d\n", ans[i]);
return 0;
}
标签:contest define print can turn space str get scanf
原文地址:https://www.cnblogs.com/DuskOB/p/10575362.html