BZOJ 2588 - 树上主席树

传送门

题目分析

类似于区间型的主席树,树上的主席树也能通过前缀的和差来计算指定路径。
建立主席树时,子节点从父节点更新,表示从根节点到该子节点的路径线段树(存放节点权值),则若要提取指定路径u->v, 先计算lca, 该路径即为u + v - lca(u, v) - fa(lca(u, v)),其余就没什么难点了。

wa了半天,原来是lca写错了!!

code

#include<bits/stdc++.h>
using namespace std;

const int N = 101000, M = 101000;
int n, m, val[N], b[N], len, fa[N][25], dep[N], dfn[N], clk;
int ecnt, adj[N], go[M << 1], nxt[M << 1];
typedef long long ll;

struct node{
	int lc, rc, cnt;
}tr[N * 25];
int pool, rt[N];

namespace IO{
	inline int read(){
	    int i = 0, f = 1; char ch = getchar();
	    for(; (ch < '0' || ch > '9') && ch != '-'; ch = getchar());
	    if(ch == '-') f = -1, ch = getchar();
	    for(; ch >= '0' && ch <= '9'; ch = getchar()) i = (i << 3) + (i << 1) + (ch -'0');
	    return i * f;
	}

	inline void wr(int x){
		if(x < 0) putchar('-'), x = -x;
		if(x > 9) wr(x / 10);
		putchar(x % 10 + '0');
	}
}using namespace IO;

inline void addEdge(int u, int v){
	nxt[++ecnt] = adj[u], adj[u] = ecnt, go[ecnt] = v;
	nxt[++ecnt] = adj[v], adj[v] = ecnt, go[ecnt] = u;
}

inline void dfs(int u, int f){
	dep[u] = dep[f] + 1;
	dfn[u] = ++clk;
	fa[u][0] = f;
	for(int i = 1; i <= 20; i++) fa[u][i] = fa[fa[u][i - 1]][i - 1];
	for(int e = adj[u], v; e ; e = nxt[e]){
		if((v = go[e]) == f) continue;
		dfs(v, u);
	}
}

inline int lca(int x, int y){
	if(dep[x] < dep[y]) swap(x, y);
	int delta = dep[x] - dep[y];
	for(int i = 20; i >= 0; i--) if(1 << i & delta) x = fa[x][i]; 
	if(x == y) return x;
	for(int i = 20; i >= 0; i--)
		if(fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
	return fa[x][0];
}

inline void disc_init(){
	sort(b + 1, b + len + 1);
	len = unique(b + 1, b + len + 1) - b - 1;
	for(int i = 1; i <= n; i++) val[i] = lower_bound(b + 1, b + len + 1, val[i]) - b;
}

inline void insert(int x, int &y, int l, int r, int v){
	tr[y = ++pool] = tr[x];
	tr[y].cnt++;
	if(l == r) return;
	int mid = l + r >> 1;
	if(v <= mid) insert(tr[x].lc, tr[y].lc, l, mid, v);
	else insert(tr[x].rc, tr[y].rc, mid + 1, r, v);
}

inline void tree_init(int u, int f){
	insert(rt[dfn[f]], rt[dfn[u]], 1, len, val[u]);
	for(int e = adj[u], v; e; e = nxt[e]){
		if((v = go[e]) == f) continue;
		tree_init(v, u);
	}
}

inline int query(int u, int v, int w, int x, int l, int r, int k){
	if(l == r) return l;
	int delta = tr[tr[u].lc].cnt + tr[tr[v].lc].cnt - tr[tr[w].lc].cnt - tr[tr[x].lc].cnt;
	int mid = l + r >> 1;
	if(delta >= k) return query(tr[u].lc, tr[v].lc, tr[w].lc, tr[x].lc, l, mid, k);
	else return query(tr[u].rc, tr[v].rc, tr[w].rc, tr[x].rc, mid + 1, r, k - delta);
}

int main(){
	//freopen("h.in", "r", stdin);
	n = read(), m = read();
	for(int i = 1; i <= n; i++) val[i] = b[++len] = read();
	disc_init();
	for(int i = 1; i < n; i++){
		int x = read(), y = read();
		addEdge(x, y);
	}
	dfs(1, 0);
	tree_init(1, 0);
	int ans = 0;
	for(int i = 1; i <= m; i++){
		int u = read() ^ ans, v = read(), k = read();
		int LCA = lca(u, v);
		int w = rt[dfn[LCA]], x = rt[dfn[fa[LCA][0]]];
		u = rt[dfn[u]], v = rt[dfn[v]];
		
		ans = b[query(u, v, w, x, 1, len, k)];
		wr(ans);
		if(i < m) putchar('\n');
	}
	return 0;
}


posted @ 2017-10-10 17:49  CzYoL  阅读(95)  评论(0编辑  收藏  举报