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HDU 5002 Tree LCT 区间更新
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Tree
Time Limit: 1 Sec
Memory Limit: 256 MB
题目连接
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=93850#problem/F
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.
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.
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.
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.
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).
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
HINT
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).
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
HINT
题意
树上，维护连边，删除边，区间加，区间定值
询问第二大的数是什么，以及有多少个的操作
题解：
LCT模板题
主要是修改pushdown和pushup这两个地方（比较麻烦……
代码：
#include <cstdio>
#include <iostream>
using namespace std;
#define N 300010
const int inf = << ;
typedef pair<int,int>dl;
struct node
{
node *p,*ch[];
int max,rev,val,addv,min;
int max2,ti2,ti1;
int setv;
int size;
}nodes[N],*cur,*null;
int n,m,u,v,w;
node *newnode(int key)
{
cur->p=cur->ch[]=cur->ch[]=null;
cur->max=cur->min=cur->val=key;
cur->max2=cur->min;
cur->ti2=;
cur->ti1=;
cur->rev=;
cur->size=;
cur->setv=-inf;
return cur++;
}
void init()
{
null=nodes;
null->p=null->ch[]=null->ch[]=null;
null->max=null->val=null->max2=-inf;
null->min=inf;
null->ti2=;
null->ti1=;
null->addv=;
null->rev=;
null->setv=-inf;
null->size=;
cur=nodes+;
}
struct dynaminctree
{
bool isroot(node *x)//判根
{
return x==null || x->p->ch[]!=x && x->p->ch[]!=x;
}
void pushup(node *x)
{
x->max = x->val , x->max2 = -inf;x->ti1=;x->ti2 = ;
x->size=; if(x->ch[] != null)
{
x->size += x->ch[]->size;
int mv = x->ch[]->max;
if(mv > x->max)
{
x->max2 = x->max;
x->ti2 = x->ti1;
x->max = mv;
x->ti1 = x->ch[]->ti1;
}
else if(mv == x->max)
{
x->ti1 += x->ch[]->ti1;
}
else if(mv > x->max2)
{
x->max2 = mv;
x->ti2 = x->ch[]->ti1;
}
else if(mv == x->max2)
{
x->ti2 += x->ch[]->ti1;
} mv = x->ch[]->max2;
if(mv != -inf)
{
if(mv > x->max2)
{
x->max2 = mv;
x->ti2 = x->ch[]->ti2;
}
else if(mv == x->max2)
{
x->ti2 += x->ch[]->ti2;
}
}
} //*****************//
if(x->ch[] != null)
{
x->size += x->ch[]->size;
int mv = x->ch[]->max;
if(mv > x->max)
{
x->max2 = x->max;
x->ti2 = x->ti1;
x->max = mv;
x->ti1 = x->ch[]->ti1;
}
else if(mv == x->max)
{
x->ti1 += x->ch[]->ti1;
}
else if(mv > x->max2)
{
x->max2 = mv;
x->ti2 = x->ch[]->ti1;
}
else if(mv == x->max2)
{
x->ti2 += x->ch[]->ti1;
} mv = x->ch[]->max2;
if(mv != -inf)
{
if(mv > x->max2)
{
x->max2 = mv;
x->ti2 = x->ch[]->ti2;
}
else if(mv == x->max2)
{
x->ti2 += x->ch[]->ti2;
}
}
} }
void pushdown(node *x)
{
if(x==null) return;
if(x->rev)
{
x->rev=;
if(x->ch[]!=null) x->ch[]->rev^=;
if(x->ch[]!=null) x->ch[]->rev^=;
swap(x->ch[],x->ch[]);
}
if(x->setv != -inf)
{
if(x->ch[] != null)
{
x->ch[]->max = x->setv , x->ch[]->ti1 = x->ch[]->size;
x->ch[]->max2 = -inf , x->ch[]->ti2 = ;
x->ch[]->setv = x->setv;
x->ch[]->addv = ;
x->ch[]->val = x->setv;
} if(x->ch[] != null)
{
x->ch[]->max = x->setv , x->ch[]->ti1 = x->ch[]->size;
x->ch[]->max2 = -inf , x->ch[]->ti2 = ;
x->ch[]->setv = x->setv;
x->ch[]->addv = ;
x->ch[]->val = x->setv;
}
x->setv = -inf;
} if(x->addv)
{
if(x->ch[] != null)
{
x->ch[]->max += x->addv;
if(x->ch[]->max2 != -inf) x->ch[]->max2 += x->addv;
x->ch[]->addv += x->addv;
x->ch[]->val += x->addv;
} if(x->ch[] != null)
{
x->ch[]->max += x->addv;
if(x->ch[]->max2 != -inf) x->ch[]->max2 += x->addv;
x->ch[]->addv += x->addv;
x->ch[]->val += x->addv;
} x->addv = ;
}
}
void rotate(node *x,int f)
{
if(isroot(x)) return;
node *y=x->p;
y->ch[!f]=x->ch[f];
x->p=y->p;
if(x->ch[f]!=null) x->ch[f]->p=y;
if(y!=null)
{
if(y==y->p->ch[]) y->p->ch[]=x;
else if(y==y->p->ch[]) y->p->ch[]=x;
}
x->ch[f]=y;
y->p=x;
pushup(y);
}
void splay(node *x)
{
static node *sta[N];
int top=;
sta[]=x;
for(node *y=x;!isroot(y);y=y->p)
sta[top++]=y->p;
while (top) pushdown(sta[--top]);
while (!isroot(x))
{
node *y=x->p;
if(isroot(y)) rotate(x,x==y->ch[]);
else
{
int f=y->p->ch[]==y;
if(y->ch[f]==x) rotate(x,!f);
else rotate(y,f);
rotate(x,f);
}
}
pushup(x);
}
node *access(node *u)
{
node *v=null;
while (u!=null)
{
splay(u);
v->p=u;
u->ch[]=v;
pushup(u);
v=u;
u=u->p;
}
return v;
}
node *link(node *u,node *v)//合并
{
access(u);
splay(u);
u->rev=;
u->p=v;
}
node *cut(node *u)//分离
{
access(u);
splay(u);
u->ch[]=u->ch[]->p=null;
pushup(u);
}
void changeroot(node *u)//换根
{
access(u)->rev^=;
}
node *getroot(node *u)//找根
{
access(u);
splay(u);
while (u->p!=null)
{
u=u->p;
}
splay(u);
return u;
}
bool queryuv(node *u,node *v)//判断是否在同一子树
{
while (u->p!=null) u=u->p;
while (v->p!=null) v=v->p;
return u==v;
}
}splay;
int eu[N],ev[N];
int main ()
{
//freopen("in.txt","r",stdin);
int t;scanf("%d",&t);
for(int cas=;cas<=t;cas++)
{
int q;
scanf("%d%d",&n,&q);
init();
for(int i=;i<=n;i++)
{
int a;
scanf("%d",&a);
newnode(a);
}
for(int i=;i<n;i++)
scanf("%d%d",&eu[i],&ev[i]);
for(int i=;i<n;i++)
splay.link(nodes+eu[i],nodes+ev[i]);
printf("Case #%d:\n",cas);
for(int i=;i<=q;i++)
{
scanf("%d",&u);
if(u==)
{
int x,y;
scanf("%d%d",&x,&y);
scanf("%d%d",&u,&v);
splay.changeroot(nodes+x);
splay.cut(nodes+y);
splay.link(nodes+u,nodes+v);
}
else if(u==)
{
scanf("%d%d%d",&u,&v,&w);
splay.changeroot(nodes+u);
splay.access(nodes+v);
node *q=splay.getroot(nodes+v);
q->max=w;
q->val=w;
if(q->max2!=-inf)q->max2=w;
q->setv=w;
}
else if(u==)
{
scanf("%d%d%d",&u,&v,&w);
splay.changeroot(nodes+u);
splay.access(nodes+v);
node *q=splay.getroot(nodes+v);
q->addv+=w;
q->max+=w;
if(q->max2!=-inf)q->max2+=w;
q->val+=w;
}
else
{
scanf("%d%d",&u,&v);
splay.changeroot(nodes+u);
splay.access(nodes+v);
node *q = splay.getroot(nodes+v);
if(q->max2 == -inf) printf("ALL SAME\n");
else printf("%d %d\n",q->max2,q->ti2);
}
}
}
return ;
}
