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

推荐论文:《树链剖分》:http://wenku.baidu.com/view/a088de01eff9aef8941e06c3.html

《QTREE解法的一些研究》:随便百度一下就有

思路:树链剖分,上面都讲得比较清楚了我就不讲了。对着树链剖分的伪代码写的,那个伪代码有一个错误(应该是错误吧……),询问那里应该是x = father[top[x]]。还有,在这题用线段树,点的权值记录与父节点的边的权值,那么最后的询问是要query(tid[x]+1, tid[y])

代码(3840MS):

 #include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std; const int MAXN = ;
const int MAXE = * MAXN;
const int INF = 0x7fffffff; int head[MAXN], cost[MAXN], id[MAXN];
int weight[MAXE], to[MAXE], next[MAXE];
int n, ecnt; inline void init() {
memset(head, , sizeof(head));
ecnt = ;
} inline void add_edge(int u, int v, int c) {
to[ecnt] = v; weight[ecnt] = c; next[ecnt] = head[u]; head[u] = ecnt++;
to[ecnt] = u; weight[ecnt] = c; next[ecnt] = head[v]; head[v] = ecnt++;
} int maxt[MAXN * ]; void modify(int x, int left, int right, int a, int b, int val) {
if(a <= left && right <= b) maxt[x] = val;
else {
int ll = x << , rr = ll ^ ;
int mid = (left + right) >> ;
if(a <= mid) modify(ll, left, mid, a, b, val);
if(mid < b) modify(rr, mid + , right, a, b, val);
maxt[x] = max(maxt[ll], maxt[rr]);
}
} int query(int x, int left, int right, int a, int b) {
if(a <= left && right <= b) return maxt[x];
else {
int ll = x << , rr = ll ^ ;
int mid = (left + right) >> , ret = ;
if(a <= mid) ret = max(ret, query(ll, left, mid, a, b));
if(mid < b) ret = max(ret, query(rr, mid + , right, a, b));
return ret;
}
} int size[MAXN], fa[MAXN], dep[MAXN], son[MAXN];
int tid[MAXN], top[MAXN], dfs_clock; void dfs_size(int u, int f, int depth) {
fa[u] = f; dep[u] = depth;
size[u] = ; son[u] = ;
int maxsize = ;
for(int p = head[u]; p; p = next[p]) {
int &v = to[p];
if(v == f) continue;
cost[v] = weight[p];
dfs_size(v, u, depth + );
size[u] += size[v];
if(size[v] > maxsize) {
maxsize = size[v];
son[u] = v;
}
}
} void dfs_heavy_edge(int u, int ancestor) {
tid[u] = ++dfs_clock; top[u] = ancestor;
modify(, , n, tid[u], tid[u], cost[u]);
if(son[u]) dfs_heavy_edge(son[u], ancestor);
for(int p = head[u]; p; p = next[p]) {
int &v = to[p];
if(v == fa[u] || v == son[u]) continue;
dfs_heavy_edge(v, v);
}
} int query(int x, int y) {
int ret = ;
while(top[x] != top[y]) {
if(dep[top[x]] < dep[top[y]]) swap(x, y);
ret = max(ret, query(, , n, tid[top[x]], tid[x]));
x = fa[top[x]];
}
if(dep[x] > dep[y]) swap(x, y);
ret = max(ret, query(, , n, tid[x] + , tid[y]));
return ret;
} void change(int x, int y) {
int u = to[x], v = to[x ^ ];
if(fa[u] == v) swap(u, v);
modify(, , n, tid[v], tid[v], y);
} char str[]; int main() {
int T; scanf("%d", &T);
for(int t = ; t <= T; ++t) {
scanf("%d", &n);
init();
for(int i = ; i < n; ++i) {
int u, v, c;
scanf("%d%d%d", &u, &v, &c);
id[i] = ecnt;
add_edge(u, v, c);
}
memset(maxt, , sizeof(maxt));
dfs_size(, , ); cost[] = -INF;
dfs_clock = ;
dfs_heavy_edge(, );
while(scanf("%s", str) && *str != 'D') {
int x, y;
scanf("%d%d", &x, &y);
if(*str == 'C') change(id[x], y);
else printf("%d\n", query(x, y));
}
}
}

代码(3400MS)(加了个IO优化……):

 #include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cctype>
using namespace std; const int MAXN = ;
const int MAXE = * MAXN;
const int INF = 0x7fffffff; int head[MAXN], cost[MAXN], id[MAXN];
int weight[MAXE], to[MAXE], next[MAXE];
int n, ecnt; inline void init() {
memset(head, , sizeof(head));
ecnt = ;
} inline void add_edge(int u, int v, int c) {
to[ecnt] = v; weight[ecnt] = c; next[ecnt] = head[u]; head[u] = ecnt++;
to[ecnt] = u; weight[ecnt] = c; next[ecnt] = head[v]; head[v] = ecnt++;
} int maxt[MAXN * ]; void modify(int x, int left, int right, int a, int b, int val) {
if(a <= left && right <= b) maxt[x] = val;
else {
int ll = x << , rr = ll ^ ;
int mid = (left + right) >> ;
if(a <= mid) modify(ll, left, mid, a, b, val);
if(mid < b) modify(rr, mid + , right, a, b, val);
maxt[x] = max(maxt[ll], maxt[rr]);
}
} int query(int x, int left, int right, int a, int b) {
if(a <= left && right <= b) return maxt[x];
else {
int ll = x << , rr = ll ^ ;
int mid = (left + right) >> , ret = ;
if(a <= mid) ret = max(ret, query(ll, left, mid, a, b));
if(mid < b) ret = max(ret, query(rr, mid + , right, a, b));
return ret;
}
} int size[MAXN], fa[MAXN], dep[MAXN], son[MAXN];
int tid[MAXN], top[MAXN], dfs_clock; void dfs_size(int u, int f, int depth) {
fa[u] = f; dep[u] = depth;
size[u] = ; son[u] = ;
int maxsize = ;
for(int p = head[u]; p; p = next[p]) {
int &v = to[p];
if(v == f) continue;
cost[v] = weight[p];
dfs_size(v, u, depth + );
size[u] += size[v];
if(size[v] > maxsize) {
maxsize = size[v];
son[u] = v;
}
}
} void dfs_heavy_edge(int u, int ancestor) {
tid[u] = ++dfs_clock; top[u] = ancestor;
modify(, , n, tid[u], tid[u], cost[u]);
if(son[u]) dfs_heavy_edge(son[u], ancestor);
for(int p = head[u]; p; p = next[p]) {
int &v = to[p];
if(v == fa[u] || v == son[u]) continue;
dfs_heavy_edge(v, v);
}
} int query(int x, int y) {
int ret = ;
while(top[x] != top[y]) {
if(dep[top[x]] < dep[top[y]]) swap(x, y);
ret = max(ret, query(, , n, tid[top[x]], tid[x]));
x = fa[top[x]];
}
if(dep[x] > dep[y]) swap(x, y);
ret = max(ret, query(, , n, tid[x] + , tid[y]));
return ret;
} void change(int x, int y) {
int u = to[x], v = to[x ^ ];
if(fa[u] == v) swap(u, v);
modify(, , n, tid[v], tid[v], y);
} char str[]; inline int readint() {
char c = getchar();
while(!isdigit(c)) c = getchar();
int ret = ;
while(isdigit(c)) ret = ret * + c - '', c = getchar();
return ret;
} int main() {
int T = readint();
for(int t = ; t <= T; ++t) {
n = readint();
init();
for(int i = ; i < n; ++i) {
int u = readint(), v = readint(), c = readint();
id[i] = ecnt;
add_edge(u, v, c);
}
memset(maxt, , sizeof(maxt));
dfs_size(, , ); cost[] = -INF;
dfs_clock = ;
dfs_heavy_edge(, );
while(scanf("%s", str) && *str != 'D') {
int x = readint(), y = readint();
if(*str == 'C') change(id[x], y);
else printf("%d\n", query(x, y));
}
}
}

SPOJ 375 Query on a tree(树链剖分)(QTREE)的更多相关文章

  1. spoj 375 Query on a tree (树链剖分)

    Query on a tree You are given a tree (an acyclic undirected connected graph) with N nodes, and edges ...

  2. SPOJ 375 Query on a tree 树链剖分模板

    第一次写树剖~ #include<iostream> #include<cstring> #include<cstdio> #define L(u) u<&l ...

  3. spoj 375 QTREE - Query on a tree 树链剖分

    题目链接 给一棵树, 每条边有权值, 两种操作, 一种是将一条边的权值改变, 一种是询问u到v路径上最大的边的权值. 树链剖分模板. #include <iostream> #includ ...

  4. SPOJ QTREE Query on a tree 树链剖分+线段树

    题目链接:http://www.spoj.com/problems/QTREE/en/ QTREE - Query on a tree #tree You are given a tree (an a ...

  5. SPOJ Query on a tree 树链剖分 水题

    You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, ...

  6. spoj QTREE - Query on a tree(树链剖分+线段树单点更新,区间查询)

    传送门:Problem QTREE https://www.cnblogs.com/violet-acmer/p/9711441.html 题解: 树链剖分的模板题,看代码比看文字解析理解来的快~~~ ...

  7. SPOJ QTREE Query on a tree ——树链剖分 线段树

    [题目分析] 垃圾vjudge又挂了. 树链剖分裸题. 垃圾spoj,交了好几次,基本没改动却过了. [代码](自带常数,是别人的2倍左右) #include <cstdio> #incl ...

  8. Query on a tree——树链剖分整理

    树链剖分整理 树链剖分就是把树拆成一系列链,然后用数据结构对链进行维护. 通常的剖分方法是轻重链剖分,所谓轻重链就是对于节点u的所有子结点v,size[v]最大的v与u的边是重边,其它边是轻边,其中s ...

  9. Bzoj 2588 Spoj 10628. Count on a tree(树链剖分LCA+主席树)

    2588: Spoj 10628. Count on a tree Time Limit: 12 Sec Memory Limit: 128 MB Description 给定一棵N个节点的树,每个点 ...

  10. SPOJ QTREE Query on a tree --树链剖分

    题意:给一棵树,每次更新某条边或者查询u->v路径上的边权最大值. 解法:做过上一题,这题就没太大问题了,以终点的标号作为边的标号,因为dfs只能给点分配位置,而一棵树每条树边的终点只有一个. ...

随机推荐

  1. OS_EVENT 信号量

    1.   OS_EVENT *T2sem=(OS_EVENT *)0; 这句代码的意思是 把OS_EVENT类型的一个指针T2sem赋值为0: 其中 OS_EVENT是数据类型,*代表是指针类型,(O ...

  2. css3中有关transform的问题

    Transform属性应用于元素的2D或3D转换.这个属性允许你将元素旋转,缩放,移动,倾斜等.

  3. Spring知识点小结(一)

    一.Spring的简介 1.spring是一个full-stack轻量级开源框架    2.spring的两大核心        IoC: inverse of control  控制反转:反转是对象 ...

  4. djano-模板层基础知识

    ########模板层######## 模板层其实就是templates文件夹里的html文件 其实这里的每个html不是真正意义的上html代码,只有经过模板渲染过后才算的上真正的html页面. 一 ...

  5. Emmet插件使用

    目录 Emmet插件使用 1.生成html5文档 2.header部分 3.body部分 Emmet插件使用 标签(空格分隔): php 前端 1.生成html5文档 html5:5 ! 2.head ...

  6. .net mvc里AutoMapper更为便捷的使用方法

    前言:AutoMapper的下载安装我就不多说了,网上百度一大堆.今天我就说说它的更为简单的使用,什么叫更为简单呢?按照一般的使用方法,我们首先建DTO,然后建每个对应的Profile,然后还要把每个 ...

  7. 密钥登录LINUX步骤

    1.创建目录2.创建一个文件3.给目录和文件授权4.关闭防火墙5.然后才可以登录.

  8. vue $set修改对象

    在vue开发中,当生成vue实例后,再次给数据赋值时,有时候并不会自动更新到视图上去: eg:<!DOCTYPE html> <html> <head> <m ...

  9. over开窗函数的用法

    over(partition by c1.pmid,d1.type,e1.objid  order by e1.objid ) pinum 先根据字段排序,pinum.在取第一条数据and p1.pi ...

  10. 如何在hadoop中使用外部的python程序文件

    业务场景大概是这样,我需要在公司hadoop集群上对博文进行结巴分词.我的数据是存储在hive表格中的,数据量涉及到五百万用户三个月内发的所有博文. 首先对于数据来说,很简单,在hive表格中就是两列 ...