SPOJ VJudge QTREE - Query on a tree

Query on a tree

Time Limit: 851MS

Memory Limit: 1572864KB

64bit IO Format: %lld & %llu

Submit Status

Description

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

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 using namespace std;
 struct edge{
     int u,v,w,next;
 }e[];
 struct node{
     int l,r,val;
     node *lc,*rc;
 }*root=NULL;
 int t,n,m,head[],js,p;
 int fa[],son[],dep[],top[],pos[],siz[];
 // fa存他的父亲 son存他的重孩子 dep它的深度 top他所在重链的链顶
 //pos在线段树中的位置 siz[v]以v为顶的子树的孩子数
 void memsett()
 {
     js=;p=;
     memset(head,,sizeof(head));
     memset(e,,sizeof(e));
     memset(son,,sizeof(son));
 }
 void add_edge(int u,int v,int w)
 {
     e[++js].u=u;e[js].v=v;e[js].w=w;
     e[js].next=head[u];head[u]=js;
 }
 void dfs(int u,int f,int d)
 {
     fa[u]=f;dep[u]=d;siz[u]=;
     for(int i=head[u];i;i=e[i].next)
     {
         int v=e[i].v;
         if(v!=f)
         {
             dfs(v,u,d+);
             siz[u]+=siz[v];
             if(!son[u]||siz[son[u]]<siz[v])
               son[u]=v;
         }
     }
 }
 void gettop(int u,int tp)
 {
     top[u]=tp;pos[u]=++p;
     if(!son[u]) return;
     gettop(son[u],tp);
     for(int i=head[u];i;i=e[i].next)
     {
         int v=e[i].v;
         if(v!=son[u]&&v!=fa[u])
           gettop(v,v);
     }
 }
 void build(node * &pt,int l,int r)
 {
      pt=new(node);
      pt->l=l;pt->r=r;pt->val=;
      if(l==r)
      {
          pt->lc=pt->rc=NULL;
          return ;
      }
      int mid=(l+r)/;
      build(pt->lc,l,mid);
      build(pt->rc,mid+,r);
 }
 void update(node * p,int ps,int val)
 {
     if(p->l==p->r)
     {
         p->val=val;return;
     }
     int mid=(p->l+p->r)/;
     if(ps<=mid) update(p->lc,ps,val);
     else update(p->rc,ps,val);
     p->val=max(p->lc->val,p->rc->val);
 }
 int query(node * p,int l,int r)
 {
     if(l<=p->l&&p->r<=r)return p->val;
     int mid=(p->l+p->r)/;
     int ans=;
     if(l<=mid)ans=max(ans,query(p->lc,l,r));
     if(r>mid)ans=max(ans,query(p->rc,l,r));
     return ans;
 }
 int find(int u,int v)
 {
     int tp1=top[u],tp2=top[v],ans=;
     while(tp1!=tp2)
     {
         if(dep[tp1]<dep[tp2])
         {
             swap(tp1,tp2);swap(u,v);
         }
         ans=max(ans,query(root,pos[tp1],pos[u]));
         u=fa[tp1];tp1=top[u];
     }
     if(u==v) return ans;
     if(dep[u]>dep[v]) swap(u,v);
     return max(ans,query(root,pos[u]+,pos[v]));
 }
 int main()
 {
     scanf("%d",&t);
     while(t--)
     {
         memsett();
         scanf("%d",&n);
         for(int i=;i<=n-;i++)
         {
             int x,y,z;
             scanf("%d%d%d",&x,&y,&z);
             add_edge(x,y,z);add_edge(y,x,z);
         }
         dfs(,,);
         gettop(,);
         build(root,,p);
         for(int i=;i<*n-;i+=)// 建边表时见了两遍
         {
             if(dep[e[i].v]<dep[e[i].u]) swap(e[i].v,e[i].u);// 让v在下面
             update(root,pos[e[i].v],e[i].w);
         }
         char s[];int u,v;
         while(scanf("%s",s)==)
         {
             if(s[]=='D') break;
             scanf("%d%d",&u,&v);
             if(s[]=='Q') printf("%d\n",find(u,v));// 查询从u到v所经过的最大边权
             else update(root,pos[e[u*-].v],v);// 将第u条边的权值修改为v
         }
     }
     return ;
 }

思路：树链剖分基础题