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) 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

 /* ***********************************************
 Author        :kuangbin
 Created Time  :2013-9-3 21:06:05
 File Name     :F:\2013ACM练习\专题学习\动态树-LCT\SPOJQTREE.cpp
 ************************************************ */

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <string>
 #include <math.h>
 #include <stdlib.h>
 #include <time.h>
 using namespace std;

 //对一颗树，进行两个操作：
 //1.修改边权
 //2.查询u->v路径上边权的最大值
 const int MAXN = ;
 int ch[MAXN][],pre[MAXN];
 int Max[MAXN],key[MAXN];
 bool rt[MAXN];
 void push_down(int r)
 {

 }
 void push_up(int r)
 {
     Max[r] = max(max(Max[ch[r][]],Max[ch[r][]]),key[r]);
 }
 void Rotate(int x)
 {
     int y = pre[x], kind = ch[y][]==x;
     ch[y][kind] = ch[x][!kind];
     pre[ch[y][kind]] = y;
     pre[x] = pre[y];
     pre[y] = x;
     ch[x][!kind] = y;
     if(rt[y])
         rt[y] = false, rt[x] = true;
     else
         ch[pre[x]][ch[pre[x]][]==y] = x;
     push_up(y);
 }
 void P(int r)
 {
     if(!rt[r])P(pre[r]);
     push_down(r);
 }
 void Splay(int r)
 {
     //P(r);
     while( !rt[r] )
     {
         int f = pre[r], ff = pre[f];
         if(rt[f])
             Rotate(r);
         else if( (ch[ff][]==f)==(ch[f][]==r) )
             Rotate(f), Rotate(r);
         else
             Rotate(r), Rotate(r);
     }
     push_up(r);
 }
 int Access(int x)
 {
     int y = ;
     do
     {
         Splay(x);
         rt[ch[x][]] = true, rt[ch[x][]=y] = false;
         push_up(x);
         x = pre[y=x];
     }
     while(x);
     return y;
 }
 //调用后u是原来u和v的lca,v和ch[u][1]分别存着lca的2个儿子
 //(原来u和v所在的2颗子树)
 void lca(int &u,int &v)
 {
     Access(v), v = ;
     while(u)
     {
         Splay(u);
         if(!pre[u])return;
         rt[ch[u][]] = true;
         rt[ch[u][]=v] = false;
         push_up(u);
         u = pre[v = u];
     }
 }

 void change(int u,int k)
 {
     Access(u);
     key[u] = k;
     push_up(u);
 }
 void query(int u,int v)
 {
     lca(u,v);
     printf("%d\n",max(Max[v],Max[ch[u][]]));
 }

 struct Edge
 {
     int to,next;
     int val;
     int index;
 }edge[MAXN*];
 int head[MAXN],tot;
 int id[MAXN];

 void addedge(int u,int v,int val,int index)
 {
     edge[tot].to = v;
     edge[tot].next = head[u];
     edge[tot].val = val;
     edge[tot].index = index;
     head[u] = tot++;
 }
 void dfs(int u)
 {
     for(int i = head[u];i != -;i = edge[i].next)
     {
         int v = edge[i].to;
         if(pre[v] != )continue;
         pre[v] = u;
         id[edge[i].index] = v;
         key[v] = edge[i].val;
         dfs(v);
     }
 }
 void init()
 {
     tot = ;
     memset(head,-,sizeof(head));
 }
 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int T;
     int n;
     int u,v,w;
     char op[];
     scanf("%d",&T);
     while(T--)
     {
         init();
         scanf("%d",&n);
         for(int i = ;i <= n;i++)
         {
             pre[i] = ;
             ch[i][] = ch[i][] = ;
             rt[i] = true;
         }
         Max[] = -;
         for(int i = ;i < n;i++)
         {
             scanf("%d%d%d",&u,&v,&w);
             addedge(u,v,w,i);
             addedge(v,u,w,i);
         }
         pre[] = -;
         dfs();
         pre[] = ;
         while(scanf("%s",&op) == )
         {
             if(op[] == 'D')break;
             scanf("%d%d",&u,&v);
             if(op[] == 'C')
                 change(id[u],v);
             else query(u,v);
         }
     }
     return ;
 }