9.27 不正常国家 Trie + 启发式合并

题意:

  每个点有一个点权， 每个点u定义一个繁忙程度， 为u子树里面两点之间的路径异或和最大的两点的异或和. 这两点的lca必须要是u. 输出每个点的繁忙程度.

题解

   x, y的路径异或和就是x到根的异或和 异或 y到根的异或和 再异或lca的值.

对于异或和01trie不太懂得:Click Here

这道题烦就烦在必须要是子树里面的两点， 这样普通的trie或者可持久trie都不可以. 那么就暴力的采用启发式合并， 对于点u， 递归建好某子树的01trie， 然后另一子树暴力的每个去查询最大异或和. 然后合并就像线段树合并一样合并， 因为01trie只有0， 1两个儿子， 如果把size大的往size小的一个一个插进去会慢.  询问把小的往大的询问优化时间. 这就是大概的思想.

#include<stdio.h>#include<cmath>#include<cstring>#include<algorithm>using namespace std;const int maxn = 130005;const int maxbit = 30;inline const int read(){      register int f = 1, x = 0;    register char ch = getchar();     while(ch < '0' || ch > '9'){if(ch == '-') f = -1;ch = getchar();}    while(ch >= '0' && ch <= '9'){x = (x << 3) + (x << 1)+ch-'0';ch = getchar();}    return f * x;}int id, Whereever_I_Go, _wanna, ed, n, m, num, top, as;   int a[maxn], c[maxn * 31][2], ans[maxn], tr[maxn], w[maxn * 31], h[maxn], s[maxn * 31], dis[maxn], siz[maxn * 31];struct edge{ int nxt, v;}e[maxn * 2];inline void add(int u, int v){e[++num].v = v, e[num].nxt = h[u], h[u] = num;e[++num].v = u, e[num].nxt = h[v], h[v] = num;}void insert(int &rt, int who, int pos){if(!rt) rt = ++id;if(pos == -1) return;Whereever_I_Go = (who >> pos) & 1;insert(c[rt][Whereever_I_Go], who, pos - 1);if(!pos) w[rt] = who;}inline int query(int now, int who){ed = 0;for(int i = maxbit; i >= 0; --i){_wanna = ((who>>i) & 1) ^ 1;int p = c[now][_wanna];if(p) ed |= (1 << i), now = p;else now = c[now][_wanna ^ 1];}return ed;}inline int merge(int x, int y){if(!x) return y;if(!y) return x;c[x][0] = merge(c[x][0], c[y][0]);c[x][1] = merge(c[x][1], c[y][1]);return x;}inline void dfs1(int rt, int pos){if(!pos) {s[++top] = w[rt]; return;}if(c[rt][0]) dfs1(c[rt][0], pos - 1);if(c[rt][1]) dfs1(c[rt][1], pos - 1);}inline void solve(int u, int v){int x = u, y = v;if(siz[tr[x]] < siz[tr[y]]) swap(x, y);top = 0, dfs1(tr[y], maxbit);for(int j = 1; j <= top; ++j) ans[u] = max(ans[u], query(tr[x], s[j] ^ a[u]));merge(tr[u], tr[v]), siz[tr[u]] += siz[tr[v]];}inline void dfs(int u, int fa){insert(tr[u], dis[u], maxbit), siz[tr[u]]++;for(int i = h[u]; i; i = e[i].nxt){int v = e[i].v;if(v == fa) continue;dis[v] = dis[u] ^ a[v];dfs(v, u); solve(u, v);}}int main(){freopen("irregular.in", "r", stdin);freopen("irregular.out", "w", stdout);n = read();int x, y;for(register int i = 1; i <= n; ++i) ans[i] = a[i] = read();for(register int i = 1; i <  n; ++i) x = read(), y = read(), add(x, y); dis[1] = a[1];dfs(1, 1);for(int i = 1; i <= n; ++i) printf("%d ", ans[i]);}