zoukankan      html  css  js  c++  java
  • 动态树(Link Cut Tree) :SPOJ 375 Query on a tree

    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。
      1 #include <iostream>
      2 #include <cstdio>
      3 #include <cstring>
      4 
      5 using namespace std;
      6 const int maxn=10010;
      7 int Max[maxn],fa[maxn],ch[maxn][2],key[maxn];
      8 bool rt[maxn];
      9 
     10 void Push_up(int p)
     11 {
     12     Max[p]=max(key[p],max(Max[ch[p][0]],Max[ch[p][1]]));
     13 }
     14 
     15 void Rotate(int x)
     16 {
     17     int y=fa[x],g=fa[y],c=ch[y][1]==x;
     18     ch[y][c]=ch[x][c^1];ch[x][c^1]=y;
     19     fa[ch[y][c]]=y;fa[y]=x;fa[x]=g;
     20     if(rt[y])
     21         rt[y]=false,rt[x]=true;
     22     else
     23         ch[g][ch[g][1]==y]=x;
     24     Push_up(y);        
     25 }
     26 
     27 void Splay(int x)
     28 {
     29     for(int y=fa[x];!rt[x];Rotate(x),y=fa[x])
     30         if(!rt[y])
     31             Rotate((ch[fa[y]][1]==y)==(ch[y][1]==x)?y:x);
     32     Push_up(x);
     33 }
     34 
     35 void Access(int x)
     36 {
     37     int y=0;
     38     while(x){
     39         Splay(x);
     40         rt[ch[x][1]]=true;
     41         rt[ch[x][1]=y]=false;
     42         Push_up(x);
     43         x=fa[y=x];
     44     }
     45 }
     46 
     47 void Query(int x,int y)
     48 {
     49     Access(y),y=0;
     50     while(true)
     51     {
     52         Splay(x);
     53         if(!fa[x]){
     54             printf("%d
    ",max(Max[y],Max[ch[x][1]]));
     55             return;
     56         }
     57         rt[ch[x][1]]=true;
     58         rt[ch[x][1]=y]=false;
     59         Push_up(x);
     60         x=fa[y=x];        
     61     }
     62 }
     63 
     64 void Change(int x,int d)
     65 {
     66     Access(x);
     67     Splay(x);
     68     key[x]=d;
     69     Push_up(x);
     70 }
     71 
     72 int fir[maxn],nxt[maxn<<1],to[maxn<<1],cnt;
     73 int e[maxn][3];
     74 
     75 void addedge(int a,int b)
     76 {
     77     nxt[++cnt]=fir[a];to[cnt]=b;fir[a]=cnt;
     78 }
     79 
     80 void DFS(int node)
     81 {
     82     for(int i=fir[node];i;i=nxt[i])
     83     {
     84         if(fa[to[i]])continue;
     85         fa[to[i]]=node;
     86         DFS(to[i]);
     87     }
     88 }
     89 
     90 void Init()
     91 {
     92     cnt=0;Max[0]=-1000000000;
     93     for(int i=1;i<=10000;i++){
     94         Max[i]=fir[i]=fa[i]=0;
     95         rt[i]=1;
     96     }
     97     return;
     98 }
     99 int main()
    100 {
    101     int T,n,a,b;
    102     scanf("%d",&T);
    103     while(T--)
    104     {
    105         Init();
    106         scanf("%d",&n);
    107         for(int i=1;i<n;i++)
    108             scanf("%d%d%d",&e[i][0],&e[i][1],&e[i][2]);
    109         for(int i=1;i<n;i++)
    110             addedge(e[i][0],e[i][1]),addedge(e[i][1],e[i][0]);
    111         fa[1]=-1;
    112         DFS(1);
    113         fa[1]=0;
    114         for(int i=1;i<n;i++)
    115         {
    116             if(fa[e[i][0]]==e[i][1])
    117                 swap(e[i][0],e[i][1]);
    118             Change(e[i][1],e[i][2]);
    119         }    
    120         char op[10];
    121         while(true)
    122         {
    123             scanf("%s",op);
    124             if(!strcmp(op,"DONE"))break;
    125             else if(!strcmp(op,"QUERY")){
    126                 scanf("%d%d",&a,&b);
    127                 Query(a,b);
    128             }
    129                 
    130             else {
    131                 scanf("%d%d",&a,&b);
    132                 Change(e[a][1],b);
    133             }
    134         }    
    135     }
    136     
    137     return 0;
    138 }

    尽最大的努力,做最好的自己!
  • 相关阅读:
    缓存地图 ArcGIS ——Local compact and exploded tile cache layer for WPF API
    ArcGIS Runtime数据制作教程
    ArcGis 的测距
    图片展示上移效果
    积累
    html/css 图片展示效果
    javascript 回到顶部效果的实现
    javascript实现下拉菜单的显示与隐藏
    html/css 实现下拉菜单效果
    信息排列效果
  • 原文地址:https://www.cnblogs.com/TenderRun/p/5221811.html
Copyright © 2011-2022 走看看