P3292 [SCOI2016]幸运数字 [线性基+倍增]

线性基+倍增

// by Isaunoya
#include <bits/stdc++.h>
using namespace std;

#define rep(i, x, y) for (register int i = (x); i <= (y); ++i)
#define Rep(i, x, y) for (register int i = (x); i >= (y); --i)
using ll = long long ;

const int _ = 1 << 21;
struct I {
	char fin[_], *p1 = fin, *p2 = fin;
	inline char gc() {
		return (p1 == p2) && (p2 = (p1 = fin) + fread(fin, 1, _, stdin), p1 == p2) ? EOF : *p1++;
	}
	inline I& operator>>(int& x) {
		bool sign = 1;
		char c = 0;
		while (c < 48) ((c = gc()) == 45) && (sign = 0);
		x = (c & 15);
		while ((c = gc()) > 47) x = (x << 1) + (x << 3) + (c & 15);
		x = sign ? x : -x;
		return *this;
	}
	inline I& operator>>(ll& x) {
		bool sign = 1;
		char c = 0;
		while (c < 48) ((c = gc()) == 45) && (sign = 0);
		x = (c & 15);
		while ((c = gc()) > 47) x = (x << 1) + (x << 3) + (c & 15);
		x = sign ? x : -x;
		return *this;
	}
} in;
struct O {
	char st[100], fout[_];
	signed stk = 0, top = 0;
	inline void flush() {
		fwrite(fout, 1, top, stdout), fflush(stdout), top = 0;
	}
	inline O& operator<<(ll x) {
		if (top > (1 << 20)) flush();
		if (x < 0) fout[top++] = 45, x = -x;
		do
			st[++stk] = x % 10 ^ 48, x /= 10;
		while (x);
		while (stk) fout[top++] = st[stk--];
		return *this;
	}
	inline O& operator<<(char x) {
		fout[top++] = x;
		return *this;
	}
	inline O& operator<<(string s) {
		if (top > (1 << 20)) flush();
		for (char x : s) fout[top++] = x;
		return *this;
	}
} out;
#define pb emplace_back
#define fir first
#define sec second

template < class T > inline void cmax(T & x , const T & y) {
	(x < y) && (x = y) ;
}
template < class T > inline void cmin(T & x , const T & y) {
	(x > y) && (x = y) ;
}

int n , q ;
const int N = 2e4 + 10 ;
ll g[N] , d[N][22][65] ;
void ins(ll x , ll * p) {
	for(int i = 63 ; ~ i ; i --)
		if(x & (1ll << i)) {
			if(! p[i]) p[i] = x ;
			x ^= p[i] ;
		}
}
void mergeto(ll * a , ll * b) { for(int i = 63 ; ~ i ; i --) if(a[i]) ins(a[i] , b) ;  }
struct edge {int v , nxt ;} ; edge e[N << 1] ;
int cnt = 0 , head[N] ;
void add(int u , int v) { e[++ cnt] = { v , head[u] } ; head[u] = cnt ; }
int fa[N] , dep[N] , f[N][22] ;
void dfs(int u) {
	for(int i = head[u] ; i ; i = e[i].nxt) {
		int v = e[i].v ;
		if(v ^ fa[u]) { fa[v] = u ; dep[v] = dep[u] + 1 ; dfs(v) ; }
	}
}
ll ans[65] ;
inline int Lca(int x , int y) {
	if(dep[x] < dep[y]) swap(x , y) ;
	for(int i = 20 ; ~ i ; i --) if(dep[f[x][i]] >= dep[y]) x = f[x][i] ;
	if(x == y) return x ;
	for(int i = 20 ; ~ i ; i --) if(f[x][i] ^ f[y][i]) { x = f[x][i] ; y = f[y][i] ; }
	return f[x][0] ;
}
inline int getdis(int lca , int y) { return dep[y] - dep[lca] + 1 ; }
inline ll query(int x , int y) {
	memset(ans , 0 , sizeof(ans)) ;
	int lca = Lca(x , y) , dis = getdis(lca , x) ;
	for(int i = 20 ; ~ i ; i --) if(dis & (1 << i)) mergeto(d[x][i] , ans) , x = f[x][i] ;
	dis = getdis(lca , y) ;
	for(int i = 20 ; ~ i ; i --) if(dis & (1 << i)) mergeto(d[y][i] , ans) , y = f[y][i] ;
	ll res = 0 ; for(int i = 63 ; ~ i ; i --) cmax(res , res ^ ans[i]) ;
	return res ;
}
signed main() {
#ifdef _WIN64
	freopen("testdata.in" , "r" , stdin) ;
#endif
	in >> n >> q ;
	rep(i , 1 , n) { in >> g[i] ; }
	rep(i , 1 , n - 1) { int u , v ; in >> u >> v ; add(u , v) ; add(v , u) ; }
	dfs(1) ; rep(i , 1 , n) f[i][0] = fa[i] ; rep(i , 1 , n) ins(g[i] , d[i][0]) ;
	rep(j , 1 , 20) 
		rep(i , 1 , n) { f[i][j] = f[f[i][j - 1]][j - 1] ; mergeto(d[i][j - 1] , d[i][j]) ; mergeto(d[f[i][j - 1]][j - 1] , d[i][j]) ; }
	while(q --) { int u , v ; in >> u >> v ; out << query(u , v) << '\n' ; }
	return out.flush(), 0;
}
posted @ 2020-01-13 20:36  _Isaunoya  阅读(157)  评论(0编辑  收藏  举报