题解
最大异或和,明显是个线性基
然而还有那么多路径……那就树分治,反正点数看起来很少,就是为了让人乘上一个60的常数嘛
把一个树的点分树记录下来,然后看看询问的两个点彼此相同的最后一个父亲是谁,把这个询问挂在这个点上,计算就暴力搜索这棵树里每一个节点到重心的线性基就行了,最后再用60的常数把两个线性基合起来
代码
#include <bits/stdc++.h>
//#define ivorysi
#define MAXN 20005
#define MAXQ 200005
typedef long long int64;
typedef unsigned int u32;
using namespace std;
struct node {
int to,next;
}edge[MAXN * 2];
int head[MAXN],sumE,n,m;
int64 val[MAXN];
vector<int> aux[MAXN],qry[MAXN];
bool vis[MAXN];
int que[MAXN],qr,ql;
int siz[MAXN],son[MAXN],fa[MAXN];
int st[MAXQ],ed[MAXQ];
int64 dp[MAXN][65],ans[MAXQ];
void add(int u,int v) {
edge[++sumE].to = v;
edge[sumE].next = head[u];
head[u] = sumE;
}
void addtwo(int u,int v) {
add(u,v);add(v,u);
}
int calc_G(int u) {
que[ql = qr = 1] = u;
fa[u] = 0;
while(ql <= qr) {
int now = que[ql++];
siz[now] = 1;son[now] = 0;
for(int i = head[now] ; i ; i = edge[i].next) {
int v = edge[i].to;
if(!vis[v] && fa[now] != v) {
que[++qr] = v;
fa[v] = now;
}
}
}
int res = que[qr];
for(int i = qr ; i >= 1 ; --i) {
int now = que[i];
siz[fa[now]] += siz[now];
if(siz[now] > son[fa[now]]) son[fa[now]] = siz[now];
if(qr - siz[now] > son[now]) son[now] = qr - siz[now];
if(son[now] < son[res]) res = now;
}
return res;
}
void dfs(int u) {
int G = calc_G(u);
vis[G] = 1;
for(int i = 1 ; i <= qr ; ++i) aux[que[i]].push_back(G);
for(int i = head[G] ; i ; i = edge[i].next) {
int v = edge[i].to;
if(!vis[v]) dfs(v);
}
}
void Init() {
scanf("%d%d",&n,&m);
for(int i = 1 ; i <= n ; ++i) scanf("%lld",&val[i]);
int u,v;
for(int i = 1 ; i < n ; ++i) {
scanf("%d%d",&u,&v);
addtwo(u,v);
}
dfs(1);
for(int i = 1 ; i <= m ; ++i) {
scanf("%d%d",&st[i],&ed[i]);
if(st[i] == ed[i]) ans[i] = val[st[i]];
}
for(int i = 1 ; i <= m ; ++i) {
if(st[i] == ed[i]) continue;
int s = min(aux[st[i]].size(),aux[ed[i]].size());
bool flag = false;
for(int j = 0 ; j < s ; ++j) {
if(aux[st[i]][j] != aux[ed[i]][j]) {
qry[aux[st[i]][j - 1]].push_back(i);
flag = 1;
break;
}
}
if(!flag) qry[aux[st[i]][s - 1]].push_back(i);
}
}
void insert(int id,int64 v) {
for(int j = 60 ; j >= 0 ; --j) {
if(v >> j & 1) {
if(dp[id][j]) v ^= dp[id][j];
else {
dp[id][j] = v;
for(int k = 60 ; k > j ; --k) {
if(dp[id][k] >> j & 1) dp[id][k] ^= dp[id][j];
}
break;
}
}
}
}
void calc(int u) {
que[ql = qr = 1] = u;
fa[u] = 0;
memset(dp[u],0,sizeof(dp[u]));
while(ql <= qr) {
int now = que[ql++];
for(int i = head[now] ; i ; i = edge[i].next) {
int v = edge[i].to;
if(!vis[v] && fa[now] != v) {
fa[v] = now;
memcpy(dp[v],dp[now],sizeof(dp[now]));
insert(v,val[v]);
que[++qr] = v;
}
}
}
}
void Process(int u) {
for(auto k : aux[u]) vis[u] = 1;
calc(u);
for(auto k : qry[u]) {
memset(dp[n + 1],0,sizeof(dp[n + 1]));
insert(n + 1,val[u]);
for(int i = 0 ; i <= 60 ; ++i) {
if(dp[st[k]][i]) insert(n + 1,dp[st[k]][i]);
if(dp[ed[k]][i]) insert(n + 1,dp[ed[k]][i]);
}
for(int i = 60 ; i >= 0 ; --i) {
if(!dp[n + 1][i]) continue;
if(!(ans[k] >> i & 1)) ans[k] ^= dp[n + 1][i];
}
}
for(auto k : aux[u]) vis[u] = 0;
}
void Solve() {
memset(vis,0,sizeof(vis));
for(int i = 1 ; i <= n ; ++i) {
if(qry[i].size() != 0) {
Process(i);
}
}
for(int i = 1 ; i <= m ; ++i) {
printf("%lld
",ans[i]);
}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Init();
Solve();
}