BZOJ 4568: [Scoi2016]幸运数字 [线性基 倍增]
4568: [Scoi2016]幸运数字
题意:一颗带点权的树,求树上两点间异或值最大子集的异或值
显然要用线性基
可以用倍增的思想,维护每个点向上\(2^j\)个祖先这些点的线性基,求lca的时候合并起来就行了
复杂度\(O(nlogn60*60)\)
注意这是点权,特判x==y的情况,需要插入a[x]
还可以用点分治和树链剖分
我的代码好慢啊...但是很好写啊
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define fir first
#define sec second
const int N=2e4+5;
inline ll read() {
char c=getchar(); ll x=0, f=1;
while(c<'0' || c>'9') {if(c=='-')f=-1; c=getchar();}
while(c>='0' && c<='9') {x=x*10+c-'0'; c=getchar();}
return x*f;
}
int n, Q, u, v; ll a[N];
struct edge{int v, ne;}e[N<<1];
int cnt=1, h[N];
inline void ins(int u, int v) {
e[++cnt]=(edge){v, h[u]}; h[u]=cnt;
e[++cnt]=(edge){u, h[v]}; h[v]=cnt;
}
namespace lb{
struct meow {
ll p[62];
meow(){memset(p, 0, sizeof(p));}
ll operator [](int x) {return p[x];}
bool insert(ll val) {
for(int i=60; i>=0; i--)
if(val & (1LL<<i)) {
if(p[i]) val ^= p[i];
else {p[i]=val; break;}
}
return val>0;
}
void merge(meow &a) {
for(int i=60; i>=0; i--) if(a[i]) insert(a[i]);
}
ll getmax() {
ll ans=0;
for(int i=60; i>=0; i--) ans = max(ans, ans^p[i]);
return ans;
}
void print() {
for(int i=60; i>=0; i--) if(p[i]) printf("p %d %lld\n",i,p[i]);
puts("");
}
}b[N][17];
}using lb::b; using lb::meow;
int fa[N][17], deep[N];
void dfs(int u) {
for(int i=1; (1<<i)<=deep[u]; i++) {
fa[u][i] = fa[ fa[u][i-1] ][i-1];
b[u][i].merge(b[u][i-1]); b[u][i].merge(b[ fa[u][i-1] ][i-1]);
}
for(int i=h[u];i;i=e[i].ne)
if(e[i].v != fa[u][0]) {
fa[e[i].v][0]=u;
deep[e[i].v]=deep[u]+1;
b[e[i].v][0].insert(a[u]);
dfs(e[i].v);
}
}
meow lca(int x, int y) { //printf("lca %d %d\n",x,y);
meow now;
if(x==y) {now.insert(a[x]); return now;}
if(deep[x]<deep[y]) swap(x, y);
int bin=deep[x]-deep[y];
for(int i=0; i<15; i++)
if((1<<i)&bin) now.merge(b[x][i]), x=fa[x][i];
if(x==y) return now;
for(int i=14; i>=0; i--)
if(fa[x][i] != fa[y][i]) now.merge(b[x][i]), now.merge(b[y][i]), x=fa[x][i], y=fa[y][i];
now.merge(b[x][0]); now.insert(a[y]);
//puts("now");now.print();
return now;
}
ll Que(int x, int y) {
return lca(x, y).getmax();
}
int main() {
freopen("in","r",stdin);
n=read(); Q=read();
for(int i=1; i<=n; i++) a[i]=read(), b[i][0].insert(a[i]);
for(int i=1; i<n; i++) ins(read(), read());
dfs(1);
//for(int i=1; i<=n; i++)
// for(int j=0; (1<<j)<=deep[i]; j++) printf("hi %d %d %d\n",i,j,fa[i][j]), b[i][j].print();
for(int i=1; i<=Q; i++) printf("%lld\n", Que(read(), read()));
}
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