SPOJ QTREE 树链剖分

Posted

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了SPOJ QTREE 树链剖分相关的知识,希望对你有一定的参考价值。

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 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

题解:
  抄板子
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include<iostream>
using namespace std ;
#define mem(a) memset(a,0,sizeof(a))
#define pb push_back
#define fi first
#define se second
#define MP make_pair
typedef long long ll;

const int maxn = 100010;
const int inf = 0x3f3f3f3f;
const int mod = 1000000007;

struct sss {
  int to,next;
}edge[maxn >> 1];
int head[maxn],tot;
int top[maxn];//top[v]表示v所在的重链的顶端节点
int fa[maxn]; //父亲节点
int deep[maxn];//深度
int num[maxn];//num[v]表示以v为根的子树的节点数
int p[maxn];//p[v]表示v与其父亲节点的连边在线段树中的位置
int fp[maxn];//和p数组相反
int son[maxn];//重儿子
int pos;
void init() {
   mem(head);tot=1;mem(son);
   pos=1;
}
void add(int u,int v) {
    edge[tot].to=v;
    edge[tot].next=head[u];
    head[u]=tot++;
}
void dfs1(int u,int pre,int d) {///第一遍dfs求出fa,deep,num,son
    deep[u]=d;
    fa[u]=pre;
    num[u]=1;
    for(int i=head[u];i;i=edge[i].next) {
        int v=edge[i].to;
        if(v != pre) {
            dfs1(v,u,d+1);
            num[u] += num[v];
            if(son[u] == 0 || num[v] > num[son[u]]) son[u]=v;
        }
    }
}
void dfs2(int u,int sp) {/////第二遍dfs求出top和p
    top[u]=sp;
    if(son[u] != 0) {
        p[u] = pos++;
        fp[p[u]] = u;
        dfs2(son[u],sp);
    }
    else {
        p[u] = pos++;
        fp[p[u]] = u;
        return ;
    }
    for(int i=head[u];i;i=edge[i].next) {
        int v=edge[i].to;
        if(v != son[u] && v != fa[u]) {
            dfs2(v,v);
        }
    }
}
////线段树
struct ss{
   int l,r;
   int M;
}tr[maxn >> 2];
void build(int k,int s,int t) {
    tr[k].l=s;tr[k].r=t;
    tr[k].M=0;
    if(s==t) return ;
    int mid = (s+t) >> 1;
    build(k<<1,s,mid);
    build(k<<1|1,mid+1,t);
}
void pushup(int k) {
    tr[k].M=max(tr[k<<1].M,tr[k<<1|1].M);
}
void update(int k,int x,int c) {
    if(tr[k].l==x&&tr[k].r==x) {
        tr[k].M=c;
        return ;
    }
    int mid = (tr[k].l+tr[k].r) >> 1;
    if(x<=mid) update(k<<1,x,c);
    else update(k<<1|1,x,c);
    pushup(k);
}
int ask(int k,int s,int t) {
    if(tr[k].l==s&&tr[k].r==t) {
        return tr[k].M;
    }
    int mid = (tr[k].l+tr[k].r) >> 1;
    if(t<=mid) return ask(k<<1,s,t);
    else if(s>mid) return ask(k<<1|1,s,t);
    else return max(ask(k<<1,s,mid),ask(k<<1|1,mid+1,t));
}
int finds(int u,int v) {
    int fx = top[u],fy = top[v];
    int tmp=0;
    while(fx!=fy) {
        if(deep[fx] < deep[fy]) {
            swap(fx,fy);
            swap(u,v);
        }
        tmp = max(tmp,ask(1,p[fx],p[u]));
        u = fa[fx];
        fx = top[u];
    }
    if(u==v) return tmp;
    if(deep[u] > deep[v]) swap(u,v);
    return max(tmp,ask(1,p[son[u]],p[v]));
}
int main() {

    int T,e[maxn][3],n;
    scanf("%d",&T);
    while(T--) {
        init();
        scanf("%d",&n);
        for(int i=1;i<n;i++) {
            scanf("%d%d%d",&e[i][0],&e[i][1],&e[i][2]);
            add(e[i][0],e[i][1]);add(e[i][1],e[i][0]);
        }
        dfs1(1,0,0);//
        dfs2(1,1);//
        build(1,1,pos-1);
        for(int i=1;i<n;i++) {
            if(deep[e[i][0]] > deep[e[i][1]]) swap(e[i][0],e[i][1]);
            update(1,p[e[i][1]],e[i][2]);
        }
        char ch[10];
        int u,v;
        while(~scanf("%s",ch)) {
            if(ch[0]==D) break;
            scanf("%d%d",&u,&v);
            if(ch[0]==Q) printf("%d\n",finds(u,v));
            else update(1,p[e[u][1]],v);
        }
    }
    return 0;
}

 

以上是关于SPOJ QTREE 树链剖分的主要内容,如果未能解决你的问题,请参考以下文章

学术篇SPOJ QTREE 树链剖分

SPOJ QTREE Query on a tree ——树链剖分 线段树

SPOJ - QTREE(树链剖分+单点更新+区间最大值查询)

[spoj 375]QTREE - Query on a tree[树链剖分]

SPOJ QTREE Query on a Tree树链剖分模板题

[SPOJ]Query on a tree(树链剖分,线段树)