动态树(Link Cut Tree) ：SPOJ 375 Query on a tree

QTREE - Query on a tree

#number-theory

 

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\n",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 }

 
posted @ 2016-02-26 21:12  TenderRun  阅读(214)  评论(0)  编辑  收藏  举报