BZOJ 2588: Spoj 10628. Count on a tree( LCA + 主席树 )

Orz..跑得还挺快的#10

自从会树链剖分后LCA就没写过倍增了...

这道题用可持久化线段树..点x的线段树表示ROOT到x的这条路径上的权值线段树

-------------------------------------------------------------------------

#include<bits/stdc++.h>

#define rep(i, n) for(int i = 0; i < n; i++)

#define clr(x, c) memset(x, c, sizeof(x))

#define M(l, r) (((l) + (r)) >> 1)

#define foreach(i, x) for(__typeof(x.begin()) i = x.begin(); i != x.end(); i++)

using namespace std;

const int maxn = 200009;

int w[maxn], n, N, id[maxn];

vector<int> G[maxn];

namespace LCA {

int top[maxn], son[maxn], dep[maxn], size[maxn], fa[maxn], TOP;

void dfs(int x) {

size[x] = 1; son[x] = -1;

foreach(it, G[x]) if(*it != fa[x]) {

dep[*it] = dep[x] + 1;

fa[*it] = x;

dfs(*it);

size[x] += size[*it];

if(!~son[x] || size[*it] > size[son[x]])

son[x] = *it;

}

void DFS(int x) {

top[x] = TOP;

if(~son[x]) DFS(son[x]);

foreach(it, G[x]) if(*it != son[x] && *it != fa[x])

DFS(TOP = *it);

}

void init() {

dep[0] = 0; fa[0] = -1;

dfs(0); DFS(TOP = 0);

}

int LCA(int x, int y) {

for(; top[x] != top[y]; x = fa[top[x]])

if(dep[top[x]] < dep[top[y]]) swap(x, y);

return dep[x] < dep[y] ? x : y;

}

struct Node {

Node *l, *r;

int s;

} pool[maxn * 40], *pt = pool, *null, *root[maxn];

void init() {

null = pt++; null->s = 0;

null->l = null->r = null;

}

int v;

Node* modify(Node* t, int l, int r) {

Node* h = pt++;

h->s = t->s + 1;

if(r > l) {

int m = M(l, r);

if(v <= m) {

h->l = modify(t->l, l, m);

h->r = t->r;

} else {

h->l = t->l;

h->r = modify(t->r, m + 1, r);

}

return h;

}

void build(int x, int fa = -1, Node* p = null) {

v = w[x] + 1;

root[x] = modify(p, 1, N);

foreach(it, G[x]) if(*it != fa)

build(*it, x, root[x]);

}

inline int read() {

char c = getchar();

int ans = 0, f = 1;

for(; !isdigit(c); c = getchar())

if(c == '-') f = -1;

for(; isdigit(c); c = getchar())

ans = ans * 10 + c - '0';

return ans * f;

}

int ans = 0;

void work(bool F) {

int x = (read()^ans) - 1, y = read() - 1, k = read(), lca = LCA::LCA(x, y);

Node *X = root[x], *Y = root[y], *A = root[lca], *P = lca ? root[LCA::fa[lca]] : null;

int L = 1, R = N;

while(L < R) {

int s = X->l->s + Y->l->s - A->l->s - P->l->s, m = M(L, R);

if(s >= k) {

X = X->l; Y = Y->l; A = A->l; P = P->l;

R = m;

} else {

X = X->r; Y = Y->r; A = A->r; P = P->r;

k -= s;

L = m + 1;

}

ans = id[L - 1];

printf("%d", ans);

if(F) putchar('\n');

}

int main() {

freopen("test.in", "r", stdin);

init();

n = read();

int m = read();

rep(i, n) {

id[i] = w[i] = read();

G[i].clear();

}

sort(id, id + n);

N = unique(id, id + n) - id;

rep(i, n) w[i] = lower_bound(id, id + N, w[i]) - id;

rep(i, n - 1) {

int u = read() - 1, v = read() - 1;

G[u].push_back(v);

G[v].push_back(u);

}

LCA::init();

build(0);

while(m--) work(m);

return 0;

}

-------------------------------------------------------------------------

2588: Spoj 10628. Count on a tree

Time Limit: 12 Sec Memory Limit: 128 MB
Submit: 2675 Solved: 606
[Submit][Status][Discuss]

Description

给定一棵N个节点的树，每个点有一个权值，对于M个询问(u,v,k)，你需要回答u xor lastans和v这两个节点间第K小的点权。其中lastans是上一个询问的答案，初始为0，即第一个询问的u是明文。

Input

第一行两个整数N,M。

第二行有N个整数，其中第i个整数表示点i的权值。

后面N-1行每行两个整数(x,y)，表示点x到点y有一条边。

最后M行每行两个整数(u,v,k)，表示一组询问。

Output

M行，表示每个询问的答案。

Sample Input

8 5
105 2 9 3 8 5 7 7
1 2
1 3
1 4
3 5
3 6
3 7
4 8
2 5 1
0 5 2
10 5 3
11 5 4
110 8 2

Sample Output

2
8
9
105
7

HINT

HINT：

N,M<=100000

暴力自重。。。

Source

鸣谢seter

转载于:https://www.cnblogs.com/JSZX11556/p/4687736.html