loj#2013. 「SCOI2016」幸运数字 点分治/线性基

题目链接

loj#2013. 「SCOI2016」幸运数字

题解

和树上路径有管...点分治吧

把询问挂到点上

求出重心后，求出重心到每个点路径上的数的线性基

对于重心为lca的合并寻味，否则标记下传

对于每个询问，只需要暴力合并两个线性基即可

每个点只会被加到logn个线性基里，所以总复杂度为O(nlogn60 + q60*2)

然后我写了句memset(b,0,sizeof 0)...被卡了1h...

代码

#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm> 

inline int read() {
	int x = 0,f = 1;
	char c = getchar();
	while(c < '0' || c > '9') c = getchar();
	while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = getchar();
	return x * f;
} 

inline long long Read() {
	long long x = 0,f = 1;
	char c = getchar();
	while(c < '0' || c > '9') c = getchar();
	while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = getchar();
	return x * f;
}
#define LL long long
const int maxn = 200007;
int n,m;
long long val[maxn];
struct node {
	int v,next;
} edge[maxn << 1];
int head[maxn],num = 0;
inline void add_edge(int u,int v) {
	edge[++ num].v = v;edge[num].next = head[u],head[u] = num;
} 

struct Base {
	LL b[63];
	inline void clear() {memset(b,0,sizeof b); }
	inline void insert(LL x) {
		for(int i = 60;~i;-- i)
			if(x >> i & 1)
				if(b[i]) x ^= b[i];
				else {b[i] = x; break;}
	}
	inline void merge(const Base &x) {
		for(int i = 60;~i;-- i)
			if(x.b[i]) insert(x.b[i]);
	}
	inline LL query() {
		LL ret = 0;
		for(int i = 60;~i;-- i)
			ret = std::max(ret ^ b[i],ret);
		return ret;
	}
} base[maxn]; 

int U[maxn],V[maxn],sz[maxn],bel[maxn];
LL ans[maxn];
std::vector<int>q[maxn];
bool vis[maxn]; 

int root = 0,mt;
void get_root(int x,int fa,int tot) {
	sz[x] = 1; int mx = 0;
	for(int i = head[x];i;i = edge[i].next) {
		int v = edge[i].v;
		if(v == fa || vis[v]) continue;
		get_root(v,x,tot);
		sz[x] += sz[v];
		mx = std::max(mx,sz[v]);
	}
	mx = std::max(tot - sz[x],mx);
	if(mx < mt) root = x, mt = mx;
}
void dfs(int x,int fa,int Bel) {
	bel[x] = Bel; base[x] = base[fa]; base[x].insert(val[x]);
	for(int i = head[x];i;i = edge[i].next)
		if(edge[i].v != fa && !vis[edge[i].v])
			dfs(edge[i].v,x,Bel);
}
int tq[maxn];
void solve(int x) {
	if(!q[x].size()) return;
	mt = 20005; get_root(x,x,sz[x]);
	vis[root] = 1;
	bel[root] = root;
	base[root].clear();
	base[root].insert(val[root]);
	for(int i = head[root];i;i = edge[i].next)
		if(!vis[edge[i].v])
		dfs(edge[i].v,root,edge[i].v);
	int tot = q[x].size();
	for(int i = 0;i <= tot;++ i) tq[i] = q[x][i];
	q[x].clear();
	Base tmp;
	for(int i = 0,id;i < tot;++ i) {
		if(bel[U[id = tq[i]]] == bel[V[id]])
			q[bel[U[id]]].push_back(id);
		else
			tmp = base[U[id]],
			tmp.merge(base[V[id]]),
			ans[id] = tmp.query();
	}
	for(int i = head[root];i;i = edge[i].next)
		if(!vis[edge[i].v]) solve(edge[i].v);
}
int main() {
	//freopen("lucky1.in","r",stdin);
	n = read();m = read();
	for(int i = 1;i <= n;++ i) val[i] = Read();
	for(int u,v,i = 1;i < n;++ i) {
		u = read(),v = read();
		add_edge(u,v); add_edge(v,u);
	}
	for(int i = 1;i <= m;++ i) {
		U[i] = read(),V[i] = read();
		if(U[i] == V[i]) ans[i] = val[U[i]];
		else q[1].push_back(i);
	}
	sz[1] = n; solve(1);
	for(int i = 1;i <= m;++ i)
	printf("%lld\n",ans[i]);
	return 0;
}