Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 821 Accepted Submission(s): 179
Problem Description
There are ) that can make him happy.
Input
There are multiple test cases. The first line of input contains an integer 5.
Output
For each test case, output the number of different pairs that can make Soda happy.
Sample Input
1
3 3
1 2 3
1 2
2 3
Sample Output
6
/*
hdu 5314 动态树
problem:
给你一个树,求有多少对(u,v)使u->v路径上面的最大值减去最小值不大于limit
solve:
最开始想的是用树链剖分维护最大最小值,结果超时 卒。。
然后看别人说动态树比树链剖分快一点,于是去学习两天动态树,然后用其维护最大最小值 卒。。
感觉没什么思路 - -
后来发现别人维护一个树的size.维持树中的最大最小值的差不大于limit
所以先按照权值排序,用两个指针. 如果 当前值-最左边值(l) > limit,就将l从这个树中除去
然后将当前节点连接到树上面,即判断与其相连的点哪些在树上面(可以用个数组来判断).
大致思路如此,然后就是如何维护size. 首先动态树中的重链是一直变化的,通常我们是用splay来维护重链上面的值,
所以需要outsize来维护哪些不在重链上面的点.
而 重链 和 普通链会在ACCESS的时候发生变化,于是在其过程中维护一下.
参考:http://blog.csdn.net/u013368721/article/details/47086899
hhh-2016-08-21 09:16:05
*/
#pragma comment(linker,"/STACK:124000000,124000000")
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <vector>
#include <map>
#define lson ch[0]
#define rson ch[1]
#define ll long long
#define clr(a,b) memset(a,b,sizeof(a))
#define key_val ch[ch[root][1]][0]
using namespace std;
const int maxn = 300100;
const int INF = 1e9+10;
ll ans ;
struct Node* null;
struct Node
{
Node* ch[2] ;
Node* fa;
int Size ;
int Outsize;
int val,rev;
void newnode(int v)
{
val = v;
Size = 1 ;
Outsize = 0;
fa = ch[0] = ch[1] = null ;
rev = 0;
}
void update_rev()
{
if(this == null)
return ;
swap(ch[0],ch[1]);
rev ^= 1;
}
void push_up () {
if(this == null)
return ;
Size = ch[0]->Size + 1 + ch[1]->Size + ch[0]->Outsize + ch[1]->Outsize;
// cout << ch[0]->Size <<" " <<ch[1] <<Size <<endl;
}
void push_down()
{
if(this == null)
return ;
if(rev)
{
ch[0]->update_rev();
ch[1]->update_rev();
rev = 0;
}
}
void link_child ( Node* to , int d )
{
ch[d] = to;
to->fa = this ;
}
int isroot()
{
return fa == null || this != fa->ch[0] && this != fa->ch[1] ;
}
void down()
{
if ( !isroot () ) fa->down () ;
push_down () ;
}
void Rotate ( int d )
{
Node* f = fa ;
Node* ff = fa->fa ;
f->link_child ( ch[d] , !d ) ;
if ( !f->isroot () )
{
if ( ff->ch[0] == f ) ff->link_child ( this , 0 ) ;
else ff->link_child ( this , 1 ) ;
}
else fa = ff ;
link_child (f,d) ;
f->push_up () ;
}
void splay ()
{
down () ;
while ( !isroot () ) {
if ( fa->isroot () ) {
this == fa->ch[0] ? Rotate ( 1 ) : Rotate ( 0 ) ;
} else {
if ( fa == fa->fa->ch[0] ) {
this == fa->ch[0] ? fa->Rotate ( 1 ) : Rotate ( 0 ) ;
Rotate ( 1 ) ;
} else {
this == fa->ch[1] ? fa->Rotate ( 0 ) : Rotate ( 1 ) ;
Rotate ( 0 ) ;
}
}
}
push_up () ;
}
void access()
{
Node* now = this ;
Node* x = null ;
while ( now != null )
{
// cout <<"now:"<< now->val <<" f:" <<x->val <<endl;
now->splay () ;
now->Outsize += (now->ch[1]->Size+now->ch[1]->Outsize);
now->Outsize -= (x->Size + x->Outsize);
now->link_child ( x , 1 );
now->push_up () ;
x = now ;
now = now->fa ;
}
splay() ;
}
void make_root()
{
access();
update_rev();
}
void cut()
{
access();
ch[0]->fa = null;
ch[0] = null;
push_up();
}
void cut(Node* to)
{
make_root();
to->cut();
}
void link(Node* to)
{
make_root();
to->make_root();
fa = to;
// cout << Size <<" " << Outsize <<endl;
ans += (ll)(to->Size + to->Outsize)*(Size+Outsize);
// cout << ans <<endl;
to->Outsize += (Size + Outsize);
push_up();
}
};
Node memory_pool[maxn];
Node* now;
Node* node[maxn];
struct Edge
{
int to,next;
}edge[maxn << 2];
int head[maxn],tot;
int vis[maxn];
void Clear()
{
now = memory_pool;
now->newnode(-INF);
null = now ++;
null->Size = 0;
tot = 0;
memset(head,-1,sizeof(head));
memset(vis,0,sizeof(vis));
}
void add_edge(int u,int v)
{
edge[tot].to = v,edge[tot].next = head[u],head[u] = tot ++;
}
struct Point
{
int id,v;
}po[maxn];
bool cmp(Point a,Point b)
{
return a.v < b.v;
}
int main()
{
int T,n,cas = 1,limit;
int x,a,b;
// freopen("in.txt","r",stdin);
scanf("%d",&T);
while(T--)
{
Clear();
scanf("%d%d",&n,&limit);
ans = 0;
for(int i = 1; i <= n; i++)
{
scanf("%d",&x);
now->newnode(x);
node[i] = now++;
po[i].v = x,po[i].id = i;
}
sort(po+1,po+1+n,cmp);
for(int i = 1; i < n; i++)
{
scanf("%d%d",&a,&b);
add_edge(a,b);
add_edge(b,a);
}
int l= 1;
for(int i =1;i <= n;i++)
{
while(l <= i && po[i].v - po[l].v > limit)
{
int u = po[l].id;
for(int j = head[u];~j;j = edge[j].next)
{
int v = edge[j].to;
if(!vis[v])
continue;
node[u]->cut(node[v]);
}
vis[u] = 0;
++l;
}
int u = po[i].id;
for(int j = head[u];~j; j = edge[j].next)
{
// cout <<edge[j].to <<endl;
int v = edge[j].to;
if(!vis[v])
continue;
node[u]->link(node[v]);
}
vis[u] = 1;
}
printf("%I64d\n",ans*2);
}
return 0;
}