Query on The Trees
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 3602 Accepted Submission(s): 1587
Problem Description
We have met so many problems on the tree, so today we will have a query problem on a set of trees.
There are N nodes, each node will have a unique weight Wi. We will have four kinds of operations on it and you should solve them efficiently. Wish you have fun!
Input
There are multiple test cases in our dataset.
For each case, the first line contains only one integer N.(1 ≤ N ≤ 300000) The next N‐1 lines each contains two integers x, y which means there is an edge between them. It also means we will give you one tree initially.
The next line will contains N integers which means the weight Wi of each node. (0 ≤ Wi ≤ 3000)
The next line will contains an integer Q. (1 ≤ Q ≤ 300000) The next Q lines will start with an integer 1, 2, 3 or 4 means the kind of this operation.
1. Given two integer x, y, you should make a new edge between these two node x and y. So after this operation, two trees will be connected to a new one.
2. Given two integer x, y, you should find the tree in the tree set who contain node x, and you should make the node x be the root of this tree, and then you should cut the edge between node y and its parent. So after this operation, a tree will be separate
into two parts.
3. Given three integer w, x, y, for the x, y and all nodes between the path from x to y, you should increase their weight by w.
4. Given two integer x, y, you should check the node weights on the path between x and y, and you should output the maximum weight on it.
Output
For each query you should output the correct answer of it. If you find this query is an illegal operation, you should output ‐1.
You should output a blank line after each test case.
Sample Input
5 1 2 2 4 2 5 1 3 1 2 3 4 5 6 4 2 3 2 1 2 4 2 3 1 3 5 3 2 1 4 4 1 4
Sample Output
3 -1 7 Hint We define the illegal situation of different operations: In first operation: if node x and y belong to a same tree, we think it‘s illegal. In second operation: if x = y or x and y not belong to a same tree, we think it‘s illegal. In third operation: if x and y not belong to a same tree, we think it‘s illegal. In fourth operation: if x and y not belong to a same tree, we think it‘s illegal.
Source
The 36th ACM/ICPC Asia Regional Dalian
Site —— Online Contest
Recommend
lcy | We have carefully selected several similar problems for you: 4004 4005 4007 4001 4008
题目大意：n个点，n-1条边，每个点都有一个点权，操作1 连接 x y,操作2断开x y，操作3 （先输入的val）x到y的路径上的点权都加val，操作4，求x到y的路径上的最大点权值
ac代码
确实kuangbin的代码要快些，跑了800多，但是我就是看不太习惯，，还是用自己习惯的写法吧。。。
#include<stdio.h>
#include<string.h>
#include<queue>
#include<iostream>
#define INF 0x7fffffff
#define N 300030
#define max(a,b) (a>b?a:b)
using namespace std;
int vis[N];
struct LCT
{
int bef[N],pre[N],next[N][2],key[N],add[N];
int rev[N],maxn[N];
void init()
{
memset(pre,0,sizeof(pre));
memset(next,0,sizeof(next));
rev[0]=rev[1]=0;
add[0]=add[1]=0;
bef[0]=bef[1]=0;
maxn[0]=key[0]=0;
}
void update_add(int x,int val)
{
if(x)
{
add[x]+=val;
key[x]+=val;
maxn[x]+=val;
}
}
void update_rev(int x)
{
if(!x)
return;
swap(next[x][0],next[x][1]);
rev[x]^=1;
}
void pushup(int x)
{
maxn[x] = max(key[x], max(maxn[next[x][0]], maxn[next[x][1]]));
}
void pushdown(int x)
{
if(add[x])
{
update_add(next[x][0],add[x]);
update_add(next[x][1],add[x]);
add[x]=0;
}
if(rev[x])
{
update_rev(next[x][0]);
update_rev(next[x][1]);
// swap(next[x][0],next[x][1]);
rev[x]=0;
}
}
void rotate(int x,int kind)
{
int y,z;
y=pre[x];
z=pre[y];
pushdown(y);
pushdown(x);
next[y][!kind]=next[x][kind];
pre[next[x][kind]]=y;
next[z][next[z][1]==y]=x;
pre[x]=z;
next[x][kind]=y;
pre[y]=x;
pushup(y);
}
void splay(int x)
{
int rt;
for(rt=x;pre[rt];rt=pre[rt]);
if(x!=rt)
{
bef[x]=bef[rt];
bef[rt]=0;
pushdown(x);
while(pre[x])
{
if(next[pre[x]][0]==x)
{
rotate(x,1);
}
else
rotate(x,0);
}
pushup(x);
}
}
void access(int x)
{
int fa;
for(fa=0;x;x=bef[x])
{
splay(x);
pushdown(x);
pre[next[x][1]]=0;
bef[next[x][1]]=x;
next[x][1]=fa;
pre[fa]=x;
bef[fa]=0;
fa=x;
pushup(x);
}
}
int getroot(int x)
{
access(x);
splay(x);
while(next[x][0])
x=next[x][0];
return x;
}
void makeroot(int x)
{
access(x);
splay(x);
update_rev(x);
}
void link(int x,int y)
{
makeroot(x);
makeroot(y);
bef[x]=y;
}
void cut(int y,int x)
{
makeroot(y);
access(x);
splay(x);
bef[next[x][0]]=bef[x];
bef[x]=0;
pre[next[x][0]]=0;
next[x][0]=0;
pushup(x);
}
void change(int x,int y,int val)
{
access(y);
for(y=0;x;x=bef[x])
{
splay(x);
if(!bef[x])
{
key[x]+=val;
update_add(y,val);
update_add(next[x][1],val);
return;
}
pushdown(x);
pre[next[x][1]]=0;
bef[next[x][1]]=x;
next[x][1]=y;
pre[y]=x;
bef[y]=0;
y=x;
pushup(x);
}
}
int query(int x,int y)
{
access(y);
for(y=0;x;x=bef[x])
{
splay(x);
if(!bef[x])
{
return max(key[x],max(maxn[next[x][1]],maxn[y]));
}
pushdown(x);
pre[next[x][1]]=0;
bef[next[x][1]]=x;
next[x][1]=y;
pre[y]=x;
bef[y]=0;
y=x;
pushup(x);
}
}
}lct;
struct s
{
int u,v,next;
}edge[N<<1];
int head[N],cnt;
void add(int u,int v)
{
edge[cnt].u=u;
edge[cnt].v=v;
edge[cnt].next=head[u];
head[u]=cnt++;
}
void bfs(int u)
{
queue<int>q;
memset(vis,0,sizeof(vis));
vis[u]=1;
q.push(u);
while(!q.empty())
{
u=q.front();
q.pop();
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].v;
if(!vis[v])
{
lct.bef[v]=u;
vis[v]=1;
q.push(v);
}
}
}
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
int i;
cnt=0;
memset(head,-1,sizeof(head));
for(i=1;i<n;i++)
{
int u,v;
scanf("%d%d",&u,&v);
add(u,v);
add(v,u);
}
lct.init();
for(i=1;i<=n;i++)
{
scanf("%d",&lct.key[i]);
}
bfs(1);
int q;
scanf("%d",&q);
while(q--)
{
int op,x,y;
scanf("%d%d%d",&op,&x,&y);
if(op==1)
{
if(lct.getroot(x)==lct.getroot(y))
{
printf("-1\n");
}
else
lct.link(x,y);
}
else
if(op==2)
{
if(x==y||lct.getroot(x)!=lct.getroot(y))
{
printf("-1\n");
}
else
lct.cut(x,y);
}
else
if(op==3)
{
int z;
scanf("%d",&z);
if(lct.getroot(y)!=lct.getroot(z))
{
printf("-1\n");
}
else
lct.change(y,z,x);
}
else
{
if(lct.getroot(x)!=lct.getroot(y))
{
printf("-1\n");
}
else
printf("%d\n",lct.query(x,y));
}
}
printf("\n");
}
}
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