正题

luogu 5838

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题目大意

给你一棵树，和若干查询，每次查询一条路径上是否有点的权值为x

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解题思路

离线处理，每次将树上权值为x的点附上1的值，然后询问就是求和，查询完后清零

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代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define N 100010
using namespace std;
int n, m, x, y, w, g, gg, tot, s[N], v[N], hs[N], fa[N], ans[N], dep[N], anc[N], head[N];
struct rec
{int to, next;
}a[N<<1];
struct node
{int x, v;bool operator < (const node b) const{return x < b.x;}
}pt[N];
struct nodee
{int t, x, y, v;bool operator < (const nodee b) const{return t < b.t;}
}q[N];
struct Tree//线段树求和
{int a[N<<2];#define ls x*2#define rs x*2+1void up(int x){a[x] = a[ls] + a[rs];return;}void add(int x, int L, int R, int w, int s){if (L == w && w == R){a[x] += s;return;}int mid = L + R >> 1;if (w <= mid) add(ls, L, mid, w, s);else add(rs, mid + 1, R, w, s);up(x);}int ask(int x, int L, int R, int l, int r){if (L == l && R == r) return a[x];int mid = L + R >> 1;if (r <= mid) return ask(ls, L, mid, l, r);if (l > mid) return ask(rs, mid + 1, R, l, r);return ask(ls, L, mid, l, mid) + ask(rs, mid + 1, R, mid + 1, r);}
}T;
void add(int x, int y)
{a[++tot].to = y;a[tot].next = head[x];head[x] = tot;return; 
}
void dfs1(int x)
{s[x] = 1;for (int i = head[x]; i; i = a[i].next)if (a[i].to != fa[x]){fa[a[i].to] = x;dep[a[i].to] = dep[x] + 1;dfs1(a[i].to);s[x] += s[a[i].to];if (s[a[i].to] > s[hs[x]]) hs[x] = a[i].to;}return;
}
void dfs2(int x, int y)
{anc[x] = y;v[x] = ++w;if (hs[x]) dfs2(hs[x], y);for (int i = head[x]; i; i = a[i].next)if (a[i].to != fa[x] && a[i].to != hs[x])dfs2(a[i].to, a[i].to);
}
int ask(int x, int y)//求和
{int g = 0;while (anc[x] != anc[y]){if (dep[anc[x]] < dep[anc[y]]) swap(x, y);g += T.ask(1, 1, n, v[anc[x]], v[x]);x = fa[anc[x]];}if (dep[x] > dep[y]) swap(x, y);g += T.ask(1, 1, n, v[x], v[y]);return g; 
}
int main()
{scanf("%d%d", &n, &m);for (int i = 1; i <= n; ++i){scanf("%d", &pt[i].x);pt[i].v = i;}for (int i = 1; i < n; ++i){scanf("%d%d", &x, &y);add(x, y);add(y, x);}for (int i = 1; i <= m; ++i){scanf("%d%d%d", &q[i].x, &q[i].y, &q[i].t);q[i].v = i;}sort(pt + 1, pt + 1 + n);sort(q + 1, q + 1 + m);fa[1] = dep[1] = 1;dfs1(1);dfs2(1, 1);g = 1;gg = 1;while(g <= n){int h = g;while(pt[h].x == pt[g].x && h <= n) T.add(1, 1, n, v[pt[h].v], 1), h++;//赋值点权while(q[gg].t < pt[g].x && gg <= m) gg++;//没有该权值的点while(q[gg].t == pt[g].x && gg <= m) ans[q[gg].v] = ask(q[gg].x, q[gg].y), gg++;//查询for (int i = g; i < h; ++i)T.add(1, 1, n, v[pt[i].v], -1);g = h;}for (int i = 1; i <= m; ++i)if (ans[i]) putchar('1');else putchar('0');return 0;
}