HDU 5877 Weak Pair 大连网络赛

题意就是说给你一棵有根树，然后定义了一种有序对叫做weak pair，两个点若满足 第一个点是第二个点的祖先，而且第一个点和第二个点的权值相乘，乘积小于他给的k，那么就是weak pair，方法很简单，就是dfs，从根节点开始向下找，看这个点和他的祖先节点有多少满足情况，比赛时的想法是dfs到某个深度的时候，把之前的所有祖先的权值存在一个数组里，然后去看有多少符合条件的加在ans里，然后超时了，gg，怎么也没想到优化，后来知道有splay树这种东西，可以很快找到有多少符合条件的，但是不会呀，所以看了别人的做法，有用treap的，就用了treap的模板（之前也对这玩意不怎么熟，只知道有这个东西不会用），方法是，看祖先节点中有多少权值小于
k/当前节点权值 ，这样子就比较快了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long LL;

int n, m;
int a[100005];
int in[100005];
int head[100005];
LL k;
LL ans;

int cnt;

struct Edge
{
int from, to;
int next;
}edge[100005];

void addedge(int u, int v)
{
edge[cnt].next = head[u];
edge[cnt].from = u;
edge[cnt].to = v;
head[u] = cnt++;
}

struct Treap
{
int size;
int fix;
LL key;
Treap *ch[2];

Treap(LL key)
{
size = 1;
fix = rand();
this->key = key;
ch[0] = ch[1] = NULL;
}
int cmp(LL x) const
{
if (x == key) return -1;
return x < key ? 0 : 1;
}
void Maintain()
{
size = 1;
if (ch[0] != NULL) size += ch[0]->size;
if (ch[1] != NULL) size += ch[1]->size;
}
};

void Rotate(Treap* &t, int d)
{
Treap *k = t->ch[d ^ 1];
t->ch[d ^ 1] = k->ch[d];
k->ch[d] = t;
t->Maintain();
k->Maintain();
t = k;
}

void Insert(Treap* &t, LL x)
{
if (t == NULL) t = new Treap(x);
else
{
int d = (x < t->key ? 0 : 1);
Insert(t->ch[d], x);
if (t->ch[d]->fix > t->fix)
Rotate(t, d ^ 1);
}
t->Maintain();
}

void Delete(Treap* &t, LL x)
{
int d = t->cmp(x);
if (d == -1)
{
Treap *tmp = t;
if (t->ch[0] == NULL)
{
t = t->ch[1];
delete tmp;
tmp = NULL;
}
else if (t->ch[1] == NULL)
{
t = t->ch[0];
delete tmp;
tmp = NULL;
}
else
{
int k = (t->ch[0]->fix > t->ch[1]->fix ? 1 : 0);
Rotate(t, k);
Delete(t->ch[k], x);
}
}
else
Delete(t->ch[d], x);
if (t != NULL) t->Maintain();
}

bool Find(Treap *t, LL x)
{
while (t != NULL)
{
int d = t->cmp(x);
if (d == -1) return true;
t = t->ch[d];
}
return false;
}

int Rank(Treap *t, LL x)
{
int r;
if (t == NULL) return 0;
if (t->ch[0] == NULL) r = 0;
else r = t->ch[0]->size;
if (x == t->key) return r + 1;
if (x < t->key)
return Rank(t->ch[0], x);
return r + 1 + Rank(t->ch[1], x);
}

void DeleteTreap(Treap* &t)
{
if (t == NULL) return;
if (t->ch[0] != NULL) DeleteTreap(t->ch[0]);
if (t->ch[1] != NULL) DeleteTreap(t->ch[1]);
delete t;
t = NULL;
}

void search(int u, Treap* &t)
{
//if (head[u] == -1) return;
LL Max = 0x3f3f3f3f3f3f3f3f;
if (a[u]) Max = k / (a[u]);
ans += Rank(t, Max);

Insert(t, a[u]);

for (int i = head[u]; i != -1; i = edge[i].next)
search(edge[i].to, t);

Delete(t, a[u]);
return;
}

int main(void)
{
//freopen("test.txt", "r", stdin);
int T;
scanf("%d", &T);
while (T--)
{
ans = 0;
cnt = 0;
memset(in, 0, sizeof in);
memset(head, -1, sizeof head);
Treap *root = NULL;
scanf("%d%lld", &n, &k);
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]);
int l, r;
for (int i = 0; i < n - 1; i++)
{
scanf("%d%d", &l, &r);
addedge(l, r);
in[r]++;
}
for (int i = 1; i <= n; i++)
{
if (in[i] == 0)
{
search(i, root);
break;
}
}
DeleteTreap(root);
printf("%lld\n", ans);
}
return 0;
}