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; 	
}