HDU4757 Tree

vjudge传送


一句话题意:给一棵树,有\(m\)次操作,每一次让你从节点\(x\)到节点\(y\)的路径中选出一个点,使这个点的权值异或\(z\)最大。(数据范围\(10^5\)


如果是一个序列,那么就是01-Trie的经典应用。
现在是一棵树,就要用到可持久化Trie啦。


可持久化Trie这东西和可持久化线段树非常的像:每一层中只把有改动的孩子建成新节点,其余的节点从以前的树中复制过来。只不过孩子个数从线段树中的两个变成了字符集的大小而已。
查询的时候和线段树一样,记录这个节点的一个权值sum,如果\(sum[new]>sum[old-1]\),说明在\([old,new]\)中这个节点是存在的。于是仿照链上查找异或最大值的方法在trie上从高位到低位跑下去就好啦。


那么对于这道题来说,\(x\)\(y\)的路径拆成\([lca, x]\)\([lca,y]\)两条路径,取两者之中的最大值就是答案,

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<queue>
#include<assert.h>
#include<ctime>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
#define forE(i, x, y) for(int i = head[x], y; ~i && (y = e[i].to); i = e[i].nxt)
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int N = 17;
const int maxn = 1e5 + 5;
const int maxN = 4e6 + 5;
In ll read()
{
	ll ans = 0;
	char ch = getchar(), las = ' ';
	while(!isdigit(ch)) las = ch, ch = getchar();
	while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
	if(las == '-') ans = -ans;
	return ans;
}
In void write(ll x)
{
	if(x < 0) x = -x, putchar('-');
	if(x >= 10) write(x / 10);
	putchar(x % 10 + '0');
}
In void MYFILE()
{
#ifndef mrclr
	freopen(".in", "r", stdin);
	freopen(".out", "w", stdout);
#endif
}

int n, m, a[maxn];
struct Edge
{
	int nxt, to;
}e[maxn << 1];
int head[maxn], ecnt = -1;
In void addEdge(int x, int y)
{
	e[++ecnt] = (Edge){head[x], y};
	head[x] = ecnt;
}

struct Trie
{
	int ch[2], sum;
}t[maxN];
int root[maxn], cnt = 0;
In void build(int old, int& now, int x, int d)
{
	t[now = ++cnt] = t[old];
	t[now].sum++;
	if(d < 0) return;
	int o = (x >> d) & 1;	
	build(t[old].ch[o], t[now].ch[o], x, d - 1);
}
In void query(int old, int now, int x, int d, int& ans)
{
	if(d < 0) return;
	int o = ((x >> d) & 1) ^ 1, Sum = t[t[now].ch[o]].sum - t[t[old].ch[o]].sum;
	if(Sum > 0) ans += (1 << d), query(t[old].ch[o], t[now].ch[o], x, d - 1, ans);
	else query(t[old].ch[o ^ 1], t[now].ch[o ^ 1], x, d - 1, ans);
}

int fa[N + 2][maxn], dep[maxn];
In void dfs(int now, int _f)
{
	build(root[_f], root[now], a[now], N);
	for(int i = 1; (1 << i) <= dep[now]; ++i)
		fa[i][now] = fa[i - 1][fa[i - 1][now]];
	forE(i, now, v)
	{
		if(v == _f) continue;
		dep[v] = dep[now] + 1, fa[0][v] = now;
		dfs(v, now);
	}
}
In int lca(int x, int y)
{
	if(dep[x] < dep[y]) swap(x, y);
	for(int i = N; i >= 0; --i)
		if(dep[fa[i][x]] >= dep[y]) x = fa[i][x];
	if(x == y) return x;
	for(int i = N; i >= 0; --i)
		if(fa[i][x] ^ fa[i][y]) x = fa[i][x], y = fa[i][y];
	return fa[0][x];
}

In void Clear()
{
	Mem(head, -1), ecnt = -1;
	Mem(fa, 0);				//一定要把倍增的数组清零…… 
	for(int i = 0; i <= cnt; ++i)
		t[i].ch[0] = t[i].ch[1] = t[i].sum = 0;
	Mem(root, 0), cnt = 0;
}

int main()
{
//	MYFILE();
	Mem(head, -1), ecnt = -1; 
	while(scanf("%d%d", &n, &m) != EOF)
	{
		for(int i = 1; i <= n; ++i) a[i] = read();
		for(int i = 1; i < n; ++i)
		{
			int x = read(), y = read();
			addEdge(x, y), addEdge(y, x); 
		}
		dep[1] = 1, dfs(1, 0);
		for(int i = 1; i <= m; ++i)
		{
			int x = read(), y = read(), p = read();
			int z = lca(x, y), ans1 = 0, ans2 = 0;
			query(root[fa[0][z]], root[x], p, N, ans1), query(root[fa[0][z]], root[y], p, N, ans2);
			write(max(ans1, ans2)), enter;
		}
		Clear();
	}
	return 0;
}
posted @ 2021-01-02 15:25  mrclr  阅读(86)  评论(0编辑  收藏  举报