[题目链接]
https://www.lydsy.com/JudgeOnline/problem.php?id=2588
[算法]
如果我们能知道“u到v这条路径上权值<= k的数的个数” , 那么就可以通过二分的方式求出答案
进一步地 ， u到v路径上权值<= k的数的个数 = u到根节点路径上权值<= k的数的个数 + v到根节点路径上权值<= k的数的个数 - u和v的最近公共祖先到根节点路径上权值<= k的数的个数 - u和v的最近公共祖先的父节点到根节点路径上权值<= k的数的个数
建立可持久化线段树 ， 查询时在线段树上二分即可
时间复杂度 : O(NlogN)
[代码]
#include<bits/stdc++.h> using namespace std; #define MAXN 100010 #define MAXLOG 20 typedef long long ll; typedef long double ld; struct edge { int to , nxt; } e[MAXN << 1]; int n , q , len , idx , tot; int lson[MAXN * 20] , rson[MAXN * 20] , sum[MAXN * 20] , depth[MAXN] , head[MAXN] , root[MAXN] , son[MAXN] , size[MAXN] , fa[MAXN] , top[MAXN]; int a[MAXN] , tmp[MAXN]; int up[MAXN][MAXLOG]; template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); } template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); } template <typename T> inline void read(T &x) { T f = 1; x = 0; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0'; x *= f; } inline void addedge(int u , int v) { ++tot; e[tot] = (edge){v , head[u]}; head[u] = tot; } inline void build(int &k , int l , int r) { k = ++idx; if (l == r) return; int mid = (l + r) >> 1; build(lson[k] , l , mid); build(rson[k] , mid + 1 , r); } inline void modify(int &k , int old , int l , int r , int pos , int value) { k = ++idx; lson[k] = lson[old] , rson[k] = rson[old]; sum[k] = sum[old] + value; if (l == r) return; int mid = (l + r) >> 1; if (mid >= pos) modify(lson[k] , lson[k] , l , mid , pos , value); else modify(rson[k] , rson[k] , mid + 1 , r , pos , value); } inline int query(int rt1 , int rt2 , int rt3 , int rt4 , int l , int r , int k) { if (l == r) return l; int value = sum[lson[rt1]] + sum[lson[rt2]] - sum[lson[rt3]] - sum[lson[rt4]]; int mid = (l + r) >> 1; if (value >= k) return query(lson[rt1] , lson[rt2] , lson[rt3] , lson[rt4] , l , mid , k); else return query(rson[rt1] , rson[rt2] , rson[rt3] , rson[rt4] , mid + 1 , r , k - value); } inline void dfs1(int u , int father) { size[u] = 1; depth[u] = depth[father] + 1; modify(root[u] , root[father] , 1 , len , a[u] , 1); for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (v == father) continue; dfs1(v , u); fa[v] = u; size[u] += size[v]; if (son[u] == 0 || size[v] > size[son[u]]) son[u] = v; } } inline void dfs2(int u , int tp) { top[u] = tp; if (son[u]) dfs2(son[u] , tp); for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (v != son[u] && v != fa[u]) dfs2(v , v); } } inline int lca(int x , int y) { while (top[x] != top[y]) { if (depth[top[x]] > depth[top[y]]) swap(x , y); y = fa[top[y]]; } if (depth[x] < depth[y]) return x; else return y; } inline int getans(int u , int v , int k) { int t1 = lca(u , v) , t2 = fa[t1]; return query(root[u] , root[v] , root[t1] , root[t2] , 1 , len , k); } int main() { read(n); read(q); for (int i = 1; i <= n; i++) { read(a[i]); tmp[i] = a[i]; } for (int i = 1; i < n; i++) { int x , y; read(x); read(y); addedge(x , y); addedge(y , x); } sort(tmp + 1 , tmp + n + 1); len = unique(tmp + 1 , tmp + n + 1) - tmp - 1; for (int i = 1; i <= n; i++) a[i] = lower_bound(tmp + 1 , tmp + len + 1 , a[i]) - tmp; build(root[0] , 1 , len); dfs1(1 , 0); dfs2(1 , 1); int lastans = 0; while (q--) { int u , v , k; read(u); read(v); read(k); u ^= lastans; printf("%d\n" , lastans = tmp[getans(u , v , k)]); } return 0; }