BZOJ 2588 Count on a tree

題面

求樹上兩點之間路徑上第k大的點權。

題解

一開始看到這道題覺得是樹剖，然后又聽說是主席樹，然后以為是主席樹+樹剖，差點沒嚇死……

然后發現，如果每個點都掛一棵主席樹，每棵都通過修改父親的主席樹得到，這樣當詢問路徑(u, v)時，u的主席樹+v的主席樹-lca的主席樹-fa[lca]的主席樹就得到了路徑上的主席樹。

#include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #define space putchar(' ') #define enter putchar('\n') using namespace std; typedef long long ll; template <class T> void read(T &x){char c;bool op = 0;while(c = getchar(), c < '0' || c > '9')if(c == '-') op = 1;x = c - '0';while(c = getchar(), c >= '0' && c <= '9')x = x * 10 + c - '0';if(op) x = -x; } template <class T> void write(T x){if(x < 0) putchar('-'), x = -x;if(x >= 10) write(x / 10);putchar('0' + x % 10); }const int N = 100005, M = 3000005; int n, m, a[N], lst[N], tot, ans; int ecnt, adj[N], nxt[2*N], go[2*N]; int fa[N], dep[N], lg[2*N], seq[2*N], cnt, pos[N], mi[2*N][20]; int idx, root[N], data[M], ls[M], rs[M];void add(int u, int v){go[++ecnt] = v;nxt[ecnt] = adj[u];adj[u] = ecnt; } void dfs(int u, int pre){fa[u] = pre, dep[u] = dep[pre] + 1;seq[++cnt] = u, pos[u] = cnt;for(int e = adj[u], v; e; e = nxt[e])if(v = go[e], v != pre)dfs(v, u), seq[++cnt] = u; } int Min(int a, int b){return dep[a] < dep[b] ? a : b; } void init(){for(int i = 1, j = 0; i <= cnt; i++){lg[i] = i == (1 << (j + 1)) ? ++j : j;mi[i][0] = seq[i];}for(int j = 1; (1 << j) <= cnt; j++)for(int i = 1; i + (1 << j) - 1 <= cnt; i++)mi[i][j] = Min(mi[i][j - 1], mi[i + (1 << (j - 1))][j - 1]); } int getlca(int u, int v){int l = pos[u], r = pos[v];if(l > r) swap(l, r);int j = lg[r - l + 1];return Min(mi[l][j], mi[r - (1 << j) + 1][j]); } void build(int &k, int l, int r){k = ++idx;if(l == r) return;int mid = (l + r) >> 1;build(ls[k], l, mid);build(rs[k], mid + 1, r); } void change(int old, int &k, int l, int r, int p){k = ++idx;data[k] = data[old] + 1, ls[k] = ls[old], rs[k] = rs[old];if(l == r) return;int mid = (l + r) >> 1;if(p <= mid) change(ls[old], ls[k], l, mid, p);else change(rs[old], rs[k], mid + 1, r, p); } int query(int u, int v, int x){int lca = getlca(u, v), l = 1, r = n, mid, sum;int k[4] = {root[u], root[v], root[lca], root[fa[lca]]};while(l < r){mid = (l + r) >> 1, sum = data[ls[k[0]]] + data[ls[k[1]]] - data[ls[k[2]]] - data[ls[k[3]]];if(x <= sum){r = mid;for(int i = 0; i < 4; i++) k[i] = ls[k[i]];}else{l = mid + 1, x -= sum;for(int i = 0; i < 4; i++) k[i] = rs[k[i]];}}return lst[l]; } void build_tree(){build(root[0], 1, tot);static int que[N], qr;que[qr = 1] = 1;for(int ql = 1; ql <= qr; ql++){int u = que[ql];change(root[fa[u]], root[u], 1, n, a[u]);for(int e = adj[u], v; e; e = nxt[e])if(v = go[e], v != fa[u])que[++qr] = v;} } int main(){read(n), read(m);for(int i = 1; i <= n; i++)read(a[i]), lst[i] = a[i];sort(lst + 1, lst + n + 1);tot = unique(lst + 1, lst + n + 1) - lst - 1;for(int i = 1; i <= n; i++)a[i] = lower_bound(lst + 1, lst + tot + 1, a[i]) - lst;for(int i = 1, u, v; i < n; i++)read(u), read(v), add(u, v), add(v, u);dfs(1, 0);init();build_tree();while(m--){int u, v, x;read(u), read(v), read(x);u ^= ans;ans = query(u, v, x);write(ans);if(m) enter;}return 0; }

轉載于:https://www.cnblogs.com/RabbitHu/p/BZOJ2588.html

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