Tree
Time Limit: 16000/8000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 920 Accepted Submission(s): 388
Problem Description
You are given a tree with N nodes which are numbered by integers 1..N. Each node is associated with an integer as the weight.
Your task is to deal with M operations of 4 types:
1.Delete an edge (x, y) from the tree, and then add a new edge (a, b). We ensure that it still constitutes a tree after adding the new edge.
2.Given two nodes a and b in the tree, change the weights of all the nodes on the path connecting node a and b (including node a and b) to a particular value x.
3.Given two nodes a and b in the tree, increase the weights of all the nodes on the path connecting node a and b (including node a and b) by a particular value d.
4.Given two nodes a and b in the tree, compute the second largest weight on the path connecting node a and b (including node a and b), and the number of times this weight occurs on the path. Note that here we need the strict second largest weight. For instance, the strict second largest weight of {3, 5, 2, 5, 3} is 3.
Your task is to deal with M operations of 4 types:
1.Delete an edge (x, y) from the tree, and then add a new edge (a, b). We ensure that it still constitutes a tree after adding the new edge.
2.Given two nodes a and b in the tree, change the weights of all the nodes on the path connecting node a and b (including node a and b) to a particular value x.
3.Given two nodes a and b in the tree, increase the weights of all the nodes on the path connecting node a and b (including node a and b) by a particular value d.
4.Given two nodes a and b in the tree, compute the second largest weight on the path connecting node a and b (including node a and b), and the number of times this weight occurs on the path. Note that here we need the strict second largest weight. For instance, the strict second largest weight of {3, 5, 2, 5, 3} is 3.
Input
The first line contains an integer T (T<=3), which means there are T test cases in the input.
For each test case, the first line contains two integers N and M (N, M<=10^5). The second line contains N integers, and the i-th integer is the weight of the i-th node in the tree (their absolute values are not larger than 10^4).
In next N-1 lines, there are two integers a and b (1<=a, b<=N), which means there exists an edge connecting node a and b.
The next M lines describe the operations you have to deal with. In each line the first integer is c (1<=c<=4), which indicates the type of operation.
If c = 1, there are four integers x, y, a, b (1<= x, y, a, b <=N) after c.
If c = 2, there are three integers a, b, x (1<= a, b<=N, |x|<=10^4) after c.
If c = 3, there are three integers a, b, d (1<= a, b<=N, |d|<=10^4) after c.
If c = 4 (it is a query operation), there are two integers a, b (1<= a, b<=N) after c.
All these parameters have the same meaning as described in problem description.
For each test case, the first line contains two integers N and M (N, M<=10^5). The second line contains N integers, and the i-th integer is the weight of the i-th node in the tree (their absolute values are not larger than 10^4).
In next N-1 lines, there are two integers a and b (1<=a, b<=N), which means there exists an edge connecting node a and b.
The next M lines describe the operations you have to deal with. In each line the first integer is c (1<=c<=4), which indicates the type of operation.
If c = 1, there are four integers x, y, a, b (1<= x, y, a, b <=N) after c.
If c = 2, there are three integers a, b, x (1<= a, b<=N, |x|<=10^4) after c.
If c = 3, there are three integers a, b, d (1<= a, b<=N, |d|<=10^4) after c.
If c = 4 (it is a query operation), there are two integers a, b (1<= a, b<=N) after c.
All these parameters have the same meaning as described in problem description.
Output
For each test case, first output "Case #x:"" (x means case ID) in a separate line.
For each query operation, output two values: the second largest weight and the number of times it occurs. If the weights of nodes on that path are all the same, just output "ALL SAME" (without quotes).
For each query operation, output two values: the second largest weight and the number of times it occurs. If the weights of nodes on that path are all the same, just output "ALL SAME" (without quotes).
Sample Input
2
3 2
1 1 2
1 2
1 3
4 1 2
4 2 3
7 7
5 3 2 1 7 3 6
1 2
1 3
3 4
3 5
4 6
4 7
4 2 6
3 4 5 -1
4 5 7
1 3 4 2 4
4 3 6
2 3 6 5
4 3 6
Sample Output
Case #1:
ALL SAME
1 2
Case #2:
3 2
1 1
3 2
ALL SAME
/*
hdu 5002 (动态树lct)
problem:
给你一棵树树,主要包含四个操作:
1 x y u v:断开x,y之间的边 连接上u,v
2 x y w:将x->y之间的点权全部置为w
3 x y w:将x->y之间的点权全部加上w
4 x y:查询x->y之间第二大的
solve:
只是需要维护下第二大值,其它直接套模板
hhh-2016-08-20 17:21:29
*/
#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 = 0x3f3f3f3f;
struct Node* null;
struct Node
{
Node* ch[2] ;
Node* fa;
int Size ;
int mMax ;
int sMax ;
int max_num ;
int Max_num ;
int val ;
int add ;
int same ;
int rev;
void newnode(int v)
{
val = v ;
mMax = v ;
sMax = -INF ;
Max_num = 1 ;
max_num = 0 ;
Size = 1 ;
add = 0 ;
same = -INF ;
fa = ch[0] = ch[1] = null ;
rev = 0;
}
void update_rev()
{
if(this == null)
return ;
swap(ch[0],ch[1]);
rev ^= 1;
}
void update_add(int v)
{
if(this == null )return ;
add += v;
mMax += v,val += v;
if(sMax != -INF) sMax += v;
}
void update_same(int v)
{
if(this == null) return ;
same = v;
add = 0,val = v,mMax = v;
sMax = -INF,Max_num = Size,max_num = 0;
}
void cal(int val,int num)
{
if ( val == -INF ) return ;
if ( val < sMax ) return ;
if ( val > mMax )
{
sMax = mMax ;
max_num = Max_num ;
mMax = val ;
Max_num = num ;
}
else if ( val == mMax )
{
Max_num += num ;
}
else if ( val > sMax )
{
sMax = val ;
max_num = num ;
}
else max_num += num ;
}
void push_up () {
Size = ch[0]->Size + 1 + ch[1]->Size ;
mMax = sMax = -INF ;
max_num = Max_num = 0 ;
cal ( val , 1 ) ;
cal ( ch[0]->mMax , ch[0]->Max_num ) ;
cal ( ch[0]->sMax , ch[0]->max_num ) ;
cal ( ch[1]->mMax , ch[1]->Max_num ) ;
cal ( ch[1]->sMax , ch[1]->max_num ) ;
}
void push_down()
{
if(rev)
{
ch[0]->update_rev();
ch[1]->update_rev();
rev = 0;
}
if(same != -INF)
{
ch[0]->update_same(same);
ch[1]->update_same(same);
same = -INF;
}
if(add)
{
ch[0]->update_add(add);
ch[1]->update_add(add);
add = 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 )
{
now->splay () ;
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();
}
Node* find_root ()
{
access () ;
Node* to = this ;
while ( to->ch[0] != null )
{
to->push_down () ;
to = to->ch[0] ;
}
return to ;
}
void cut(Node* to)
{
to->make_root();
cut();
}
void link(Node* to)
{
to->make_root();
to->fa = this;
}
void make_same(Node* to,int val)
{
to->make_root();
access();
update_same(val);
}
void make_add(Node* to,int val)
{
to->make_root();
access();
update_add(val);
}
void query(Node* to)
{
to->make_root();
access();
if(!max_num)
printf("ALL SAME
");
else
printf("%d %d
",sMax,max_num);
}
};
Node memory_pool[maxn];
Node* now;
Node* node[maxn];
void Clear()
{
now = memory_pool;
now->newnode(-INF);
null = now ++;
null->Size = 0;
}
int main()
{
int T,n,cas = 1,m;
int x,y,a,b,c;
int ob;
// freopen("in.txt","r",stdin);
scanf("%d",&T);
while(T--)
{
Clear();
scanf("%d%d",&n,&m);
printf("Case #%d:
",cas++);
for(int i = 1; i <= n; i++)
{
scanf("%d",&x);
now->newnode(x);
node[i] = now++;
}
for(int i = 1; i < n; i++)
{
scanf("%d%d",&a,&b);
node[a]->link(node[b]);
}
for(int i= 1; i <= m; i++)
{
scanf("%d",&ob);
if(ob == 1)
{
scanf("%d%d%d%d",&x,&y,&a,&b);
node[x]->cut(node[y]);
node[a]->link(node[b]);
}
else if(ob == 2)
{
scanf("%d%d%d",&x,&y,&c);
node[x]->make_same(node[y],c);
}
else if(ob == 3)
{
scanf("%d%d%d",&x,&y,&c);
node[x]->make_add(node[y],c);
}
else if(ob == 4)
{
scanf("%d%d",&x,&y);
node[x]->query(node[y]);
}
}
}
return 0;
}