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  • SPOJ VJudge QTREE

    Time Limit: 851MS   Memory Limit: 1572864KB   64bit IO Format: %lld & %llu

     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 ab 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
      1 #include<iostream>
      2 #include<cstdio>
      3 #include<cstring>
      4 using namespace std;
      5 struct edge{
      6     int u,v,w,next;
      7 }e[20010];
      8 struct node{
      9     int l,r,val;
     10     node *lc,*rc;
     11 }*root=NULL;
     12 int t,n,m,head[10010],js,p;
     13 int fa[10010],son[10010],dep[10010],top[10010],pos[10010],siz[10010];
     14 // 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 
     15 //pos在线段树中的位置 siz[v]以v为顶的子树的孩子数 
     16 void memsett()
     17 {
     18     js=0;p=0;
     19     memset(head,0,sizeof(head));
     20     memset(e,0,sizeof(e));
     21     memset(son,0,sizeof(son));
     22 }
     23 void add_edge(int u,int v,int w)
     24 {
     25     e[++js].u=u;e[js].v=v;e[js].w=w;
     26     e[js].next=head[u];head[u]=js;
     27 }
     28 void dfs(int u,int f,int d)
     29 {
     30     fa[u]=f;dep[u]=d;siz[u]=1;
     31     for(int i=head[u];i;i=e[i].next)
     32     {
     33         int v=e[i].v;
     34         if(v!=f)
     35         {
     36             dfs(v,u,d+1);
     37             siz[u]+=siz[v];
     38             if(!son[u]||siz[son[u]]<siz[v])
     39               son[u]=v;
     40         }
     41     }
     42 }
     43 void gettop(int u,int tp)
     44 {
     45     top[u]=tp;pos[u]=++p;
     46     if(!son[u]) return;
     47     gettop(son[u],tp);
     48     for(int i=head[u];i;i=e[i].next)
     49     {
     50         int v=e[i].v;
     51         if(v!=son[u]&&v!=fa[u])
     52           gettop(v,v);
     53     }
     54 }
     55 void build(node * &pt,int l,int r)
     56 {
     57      pt=new(node);
     58      pt->l=l;pt->r=r;pt->val=0;
     59      if(l==r)
     60      {
     61          pt->lc=pt->rc=NULL;
     62          return ;
     63      }
     64      int mid=(l+r)/2;
     65      build(pt->lc,l,mid);
     66      build(pt->rc,mid+1,r);
     67 }
     68 void update(node * p,int ps,int val)
     69 {
     70     if(p->l==p->r)
     71     {
     72         p->val=val;return;
     73     }
     74     int mid=(p->l+p->r)/2;
     75     if(ps<=mid) update(p->lc,ps,val);
     76     else update(p->rc,ps,val);
     77     p->val=max(p->lc->val,p->rc->val);
     78 }
     79 int query(node * p,int l,int r)
     80 {
     81     if(l<=p->l&&p->r<=r)return p->val;
     82     int mid=(p->l+p->r)/2;
     83     int ans=0;
     84     if(l<=mid)ans=max(ans,query(p->lc,l,r));
     85     if(r>mid)ans=max(ans,query(p->rc,l,r));
     86     return ans;
     87 }
     88 int find(int u,int v)
     89 {
     90     int tp1=top[u],tp2=top[v],ans=0;
     91     while(tp1!=tp2)
     92     {
     93         if(dep[tp1]<dep[tp2])
     94         {
     95             swap(tp1,tp2);swap(u,v);
     96         }
     97         ans=max(ans,query(root,pos[tp1],pos[u]));
     98         u=fa[tp1];tp1=top[u];
     99     }
    100     if(u==v) return ans;
    101     if(dep[u]>dep[v]) swap(u,v);
    102     return max(ans,query(root,pos[u]+1,pos[v]));
    103 } 
    104 int main()
    105 {
    106     scanf("%d",&t);
    107     while(t--)
    108     {
    109         memsett();
    110         scanf("%d",&n);
    111         for(int i=1;i<=n-1;i++)
    112         {
    113             int x,y,z;
    114             scanf("%d%d%d",&x,&y,&z);
    115             add_edge(x,y,z);add_edge(y,x,z);
    116         }
    117         dfs(1,0,1);
    118         gettop(1,1);
    119         build(root,1,p);
    120         for(int i=1;i<2*n-2;i+=2)// 建边表时见了两遍 
    121         {
    122             if(dep[e[i].v]<dep[e[i].u]) swap(e[i].v,e[i].u);// 让v在下面 
    123             update(root,pos[e[i].v],e[i].w);
    124         }
    125         char s[15];int u,v;
    126         while(scanf("%s",s)==1)
    127         {
    128             if(s[0]=='D') break;
    129             scanf("%d%d",&u,&v);
    130             if(s[0]=='Q') printf("%d
    ",find(u,v));// 查询从u到v所经过的最大边权 
    131             else update(root,pos[e[u*2-1].v],v);// 将第u条边的权值修改为v 
    132         }
    133     }
    134     return 0;
    135 }
    思路:树链剖分基础题
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  • 原文地址:https://www.cnblogs.com/suishiguang/p/6105144.html
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