UVa 13101 - Tobby on Tree

contents

  1. 1. Problem
  2. 2. Sample Input
  3. 3. Sample Output
  4. 4. Solution

Problem

一場遊戲有 $N$ 個數,每一次挑兩個不同的數,將大的數字扣除小的數字,直到沒辦法挑出任何不同的數,亦即所有數皆相同,遊戲結果就是最後的數字值。

現在給予一棵樹,操作有兩種:

  • 詢問樹上兩點 $u$$v$ 的路徑上的數字,遊戲結果為何。
  • 改變樹上某個節點 $u$ 的值為 $c$

Sample Input

1
2
3
4
5
6
7
8
9
10
5
5 15 20 15 9
0 2
0 3
3 1
3 4
3
1 2 1
2 3 3
1 1 4

Sample Output

1
2
5
3

Solution

明顯地答案就是路徑上所有數字的最大公因數,充分利用 Link/Cut Tree 的特性,解決樹上詢問的詢問即可。

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

class LCT { // Link-Cut Tree
public:
static const int MAXN = 262144;
struct Node {
static Node *EMPTY;
Node *ch[2], *fa;
int rev;
int val, size;
int gcd;
Node() {
ch[0] = ch[1] = fa = NULL;
rev = 0;
val = 0;
gcd = 1, size = 1;
}
bool is_root() {
return fa->ch[0] != this && fa->ch[1] != this;
}
void pushdown() {
if (rev) {
ch[0]->rev ^= 1, ch[1]->rev ^= 1;
swap(ch[0], ch[1]);
rev ^= 1;
}
}
void pushup() {
if (this == EMPTY) return ;
gcd = this->val, size = 1;
if (ch[0] != EMPTY) {
gcd = __gcd(gcd, ch[0]->gcd);
size += ch[0]->size;
}
if (ch[1] != EMPTY) {
gcd = __gcd(gcd, ch[1]->gcd);
size += ch[1]->size;
}
}
} _mem[MAXN];

int bufIdx;
LCT() {
Node::EMPTY = &_mem[0];
Node::EMPTY->fa = Node::EMPTY->ch[0] = Node::EMPTY->ch[1] = Node::EMPTY;
Node::EMPTY->size = 0;
bufIdx = 1;
}
void init() {
bufIdx = 1;
}
Node* newNode() {
Node *u = &_mem[bufIdx++];
*u = Node();
u->fa = u->ch[0] = u->ch[1] = Node::EMPTY;
return u;
}
void rotate(Node *x) {
Node *y;
int d;
y = x->fa, d = y->ch[1] == x ? 1 : 0;
x->ch[d^1]->fa = y, y->ch[d] = x->ch[d^1];
x->ch[d^1] = y;
if (!y->is_root())
y->fa->ch[y->fa->ch[1] == y] = x;
x->fa = y->fa, y->fa = x;
y->pushup(), x->pushup();
}
void deal(Node *x) {
if (!x->is_root()) deal(x->fa);
x->pushdown();
}
void splay(Node *x) {
Node *y, *z;
deal(x);
while (!x->is_root()) {
y = x->fa, z = y->fa;
if (!y->is_root()) {
if (y->ch[0] == x ^ z->ch[0] == y)
rotate(x);
else
rotate(y);
}
rotate(x);
}
x->pushup();
}
Node* access(Node *u) {
Node *v = Node::EMPTY;
for (; u != Node::EMPTY; u = u->fa) {
splay(u);
u->ch[1] = v;
v = u;
v->pushup();
}
return v;
}
void mk_root(Node *u) {
access(u)->rev ^= 1, splay(u);
}
void cut(Node *x, Node *y) {
mk_root(x);
access(y), splay(y);
y->ch[0] = x->fa = Node::EMPTY;
}
Node* _cut(Node *rt, Node *x) {
Node *u, *v;
mk_root(rt);
access(x), splay(x);
for (v = x->ch[0]; v->ch[1] != Node::EMPTY; v = v->ch[1]);
x->ch[0]->fa = x->fa;
x->fa = x->ch[0] = Node::EMPTY;
return v;
}
void link(Node *x, Node *y) {
mk_root(x);
x->fa = y;
}
Node* find(Node *x) {
for (x = access(x); x->pushdown(), x->ch[0] != Node::EMPTY; x = x->ch[0]);
return x;
}
//
int gcdPath(Node *x, Node *y) {
mk_root(x);
access(y), splay(y);
return y->gcd;
}
//
void changeNode(Node *x, int c) {
mk_root(x);
x->val = c;
}
void debug(int n) {
}
} tree;
LCT::Node *LCT::Node::EMPTY;
LCT::Node *A[262144];

int W[131072];
int main() {
int n, m, cmd, x, y;
while (scanf("%d", &n) == 1) {
for (int i = 0; i < n; i++)
scanf("%d", &W[i]);
tree.init();
for (int i = 0; i < n; i++) {
A[i] = tree.newNode();
A[i]->val = W[i];
}
for (int i = 1; i < n; i++) {
scanf("%d %d", &x, &y);
tree.link(A[x], A[y]);
}
scanf("%d", &m);
for (int i = 0; i < m; i++) {
scanf("%d %d %d", &cmd, &x, &y);
if (cmd == 1) {
printf("%d\n", tree.gcdPath(A[x], A[y]));
} else {
tree.changeNode(A[x], y);
}
}
}
return 0;
}
/*
5
5 15 20 15 9
0 2
0 3
3 1
3 4
9999
1 2 1
1 1 3
2 3 3
1 1 4
1 1 3
*/