QTREE - Query on a tree

You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3...N-1.

We will ask you to perfrom some instructions of the following form:

  • CHANGE i ti : change the cost of the i-th edge to ti
    or
  • QUERY a b : ask for the maximum edge cost on the path from node a to node b

Input

The first line of input contains an integer t, the number of test cases (t <= 20). t test cases follow.

For each test case:

  • In the first line there is an integer N (N <= 10000),
  • In the next N-1 lines, the i-th line describes the i-th edge: a line with three integers a b c denotes an edge between a, b of cost c (c <= 1000000),
  • The next lines contain instructions "CHANGE i ti" or "QUERY a b",
  • The end of each test case is signified by the string "DONE".

There is one blank line between successive tests.

Output

For each "QUERY" operation, write one integer representing its result.

Example

Input:
1 3
1 2 1
2 3 2
QUERY 1 2
CHANGE 1 3
QUERY 1 2
DONE Output:
1
3   这题可以用树链剖分做,我这里用LCT做的,代码量更少。
 #include <iostream>
#include <cstdio>
#include <cstring> using namespace std;
const int maxn=;
int Max[maxn],fa[maxn],ch[maxn][],key[maxn];
bool rt[maxn]; void Push_up(int p)
{
Max[p]=max(key[p],max(Max[ch[p][]],Max[ch[p][]]));
} void Rotate(int x)
{
int y=fa[x],g=fa[y],c=ch[y][]==x;
ch[y][c]=ch[x][c^];ch[x][c^]=y;
fa[ch[y][c]]=y;fa[y]=x;fa[x]=g;
if(rt[y])
rt[y]=false,rt[x]=true;
else
ch[g][ch[g][]==y]=x;
Push_up(y);
} void Splay(int x)
{
for(int y=fa[x];!rt[x];Rotate(x),y=fa[x])
if(!rt[y])
Rotate((ch[fa[y]][]==y)==(ch[y][]==x)?y:x);
Push_up(x);
} void Access(int x)
{
int y=;
while(x){
Splay(x);
rt[ch[x][]]=true;
rt[ch[x][]=y]=false;
Push_up(x);
x=fa[y=x];
}
} void Query(int x,int y)
{
Access(y),y=;
while(true)
{
Splay(x);
if(!fa[x]){
printf("%d\n",max(Max[y],Max[ch[x][]]));
return;
}
rt[ch[x][]]=true;
rt[ch[x][]=y]=false;
Push_up(x);
x=fa[y=x];
}
} void Change(int x,int d)
{
Access(x);
Splay(x);
key[x]=d;
Push_up(x);
} int fir[maxn],nxt[maxn<<],to[maxn<<],cnt;
int e[maxn][]; void addedge(int a,int b)
{
nxt[++cnt]=fir[a];to[cnt]=b;fir[a]=cnt;
} void DFS(int node)
{
for(int i=fir[node];i;i=nxt[i])
{
if(fa[to[i]])continue;
fa[to[i]]=node;
DFS(to[i]);
}
} void Init()
{
cnt=;Max[]=-;
for(int i=;i<=;i++){
Max[i]=fir[i]=fa[i]=;
rt[i]=;
}
return;
}
int main()
{
int T,n,a,b;
scanf("%d",&T);
while(T--)
{
Init();
scanf("%d",&n);
for(int i=;i<n;i++)
scanf("%d%d%d",&e[i][],&e[i][],&e[i][]);
for(int i=;i<n;i++)
addedge(e[i][],e[i][]),addedge(e[i][],e[i][]);
fa[]=-;
DFS();
fa[]=;
for(int i=;i<n;i++)
{
if(fa[e[i][]]==e[i][])
swap(e[i][],e[i][]);
Change(e[i][],e[i][]);
}
char op[];
while(true)
{
scanf("%s",op);
if(!strcmp(op,"DONE"))break;
else if(!strcmp(op,"QUERY")){
scanf("%d%d",&a,&b);
Query(a,b);
} else {
scanf("%d%d",&a,&b);
Change(e[a][],b);
}
}
} return ;
}

动态树(Link Cut Tree) :SPOJ 375 Query on a tree的更多相关文章

  1. SPOJ 375. Query on a tree (动态树)

    375. Query on a tree Problem code: QTREE You are given a tree (an acyclic undirected connected graph ...

  2. SPOJ 375. Query on a tree (树链剖分)

    Query on a tree Time Limit: 5000ms Memory Limit: 262144KB   This problem will be judged on SPOJ. Ori ...

  3. spoj 375 Query on a tree(树链剖分,线段树)

      Query on a tree Time Limit: 851MS   Memory Limit: 1572864KB   64bit IO Format: %lld & %llu Sub ...

  4. spoj 375 Query on a tree (树链剖分)

    Query on a tree You are given a tree (an acyclic undirected connected graph) with N nodes, and edges ...

  5. SPOJ 375 Query on a tree【树链剖分】

    题目大意:给你一棵树,有两个操作1.修改一条边的值,2.询问从x到y路径上边的最大值 思路:如果树退化成一条链的话线段树就很明显了,然后这题就是套了个树连剖分,调了很久终于调出来第一个模板了 #inc ...

  6. SPOJ 375 Query on a tree 树链剖分模板

    第一次写树剖~ #include<iostream> #include<cstring> #include<cstdio> #define L(u) u<&l ...

  7. SPOJ 375 Query on a tree(树链剖分)(QTREE)

    You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, ...

  8. SPOJ 375 Query on a tree(树链剖分)

    https://vjudge.net/problem/SPOJ-QTREE 题意: 给出一棵树,树上的每一条边都有权值,现在有查询和更改操作,如果是查询,则要输出u和v之间的最大权值. 思路: 树链剖 ...

  9. spoj 375 query on a tree LCT

    这道题是树链剖分的裸题,正在学LCT,用LCT写了,发现LCT代码比树链剖分还短点(但我的LCT跑极限数据用的时间大概是kuangbin大神的树链剖分的1.6倍,所以在spoj上是850ms卡过的). ...

随机推荐

  1. Ubuntu16.04LTS安装

    1. 制作u盘启动盘 下载ubuntu-16.04-desktop-amd64.iso文件后,使用u盘启动盘制作工具:Win32DiskImager(14.04LTS后都需要用到这工具制作:https ...

  2. 19、XHTML

    XHTML 可扩展超文本标签语言(EXtensible HyperText Markup Language). 是一种 W3C 标准. 更严格,更纯净的HTML代码. 目标是取代HTML代码. XHT ...

  3. linux,安装软件报错cannot create regular file '/usr/local/man/man1': No such file or directory

    make install时报错,如下 install: cannot create regular file '/usr/local/man/man1': No such file or direct ...

  4. Oracle 左连接、右连接、全外连接、(+)号作用、inner join(等值连接) (转载)

    Oracle  外连接 (1)左外连接 (左边的表不加限制)       (2)右外连接(右边的表不加限制)       (3)全外连接(左右两表都不加限制) 外连接(Outer Join) oute ...

  5. wpf 窗体中显示当前系统时间

    先看一下效果: 这其实是我放置了两个TextBlock,上面显示当前的日期,下面显示时间. 接下来展示一下代码: 在XAML中: <StackPanel Width="205" ...

  6. SGU 231.Prime Sum

    题意: 求有多少对质数(a,b)满足a<=b 且a+b也为质数.(a+b<=10^6) Solution: 除了2之外的质数都是奇数,两个奇数的和是偶数,不可能是质数.所以题目就是求差为2 ...

  7. placeholder调整颜色

    placeholder需要设定以下样式: ::-webkit-input-placeholder { /* WebKit browsers */ color: #999; } :-moz-placeh ...

  8. C#编程连接数据库,通过更改配置文件切换数据库功能。

           该实例主要应用情景:假如某公司用mysql当做数据库服务器,由于发现mysql数据库在运行中出现不稳定情况,针对这情况,厂家要求更换连接数据库方式,改用SQL server数据库,来满足 ...

  9. Python 的“+”和append在添加字符串时候的区别

    对于一个空的Python列表,往后添加内容有很多种,其中两种一个是用“+”直接添加内容,另外一种是Listname.append(x)来添加内容 其中,如果处理字符串 在使用“+”的时候,会将字符串拆 ...

  10. Shell 控制并发

    方法1: #!/bin/bash c=0 for i in `seq -w 18 31`;do while [ $c -ge 3 ];do c=$(jobs -p |wc -w) sleep 1s d ...