SPOJ 375 Query on a tree(树链剖分)(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 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
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
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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, ...
- 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~~~ ...
- 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 ...
- 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 ...
- 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 ...
- 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
- 使用带有数组的 ng-bind
<!DOCTYPE html><html><head><meta http-equiv="Content-Type" content=&q ...
- html基础用法(下)
设计表格: <html> <head> <title>表格</title> <meta charset="utf-8" /&g ...
- CRS
CRS是集群就绪服务(cluster ready service)的简称,主要负责集群中的资源管理以及OCR管理.为了与10gR2集群管理软件名称crs区分,这里用CRSD代替CRS.相关概念:--资 ...
- Object C学习笔记20-结构体(转)
在学习Object C中的过程中,关于struct的资料貌似非常少,查阅了C方面的资料总结了一些学习心得! 一. 定义结构 结构体是一种数据类型的组合和数据抽象.结构体的定义语法如下: struct ...
- ABAP术语-Business Scenario
Business Scenario 原文:http://www.cnblogs.com/qiangsheng/archive/2008/01/12/1035980.html End-to-end co ...
- Eclipse关联tomcat
一,添加Tomcat Windows-->Preferences-->Server-->Runtime Enviroment添加一个tomcat,这里选择tomcat8.0 Next ...
- (第01节)IDEA快速搭建web项目
在配置好环境,熟悉了IDEA的基本操作后,就要开始搭建WEB项目了: File——>new——>project——>然后选择Maven 点击Create from archetype ...
- 路由器基础配置之广播多路访问链路上的ospf
我们将以上面的拓扑图进行实验,因为是要以不断广播的形式进行ospf,所有中间加了一个集线器,这种ospf和前一种不同,路由器之间会在配置好ospf之后选举出一个老大,DR,一个备份,BDR,而其他路由 ...
- vue $set修改数组
看了别人写的,自己简单写一下自己的理解. 因为 JavaScript 的限制,Vue.js 不能检测到下面数组变化,所以,想要正常是不能通过操作数组来渲染dom的,解决的方法是通过set方法, 在组件 ...
- 配置Echarts大全
由于项目中需要用到Echarts,最近研究了一个星期.网上的教程也挺多的.磕磕碰碰的,难找到合适的例子.都说的马马虎虎.不废话了.开始. 这种上下排列的... 还有这种地图的.(如下) 还有就是配置的 ...