Query on a tree 树链剖分 [模板]
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
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 100005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
#define mclr(x,a) memset((x),a,sizeof(x))
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii; inline int rd() {
int x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
} ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; } /*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
struct node {
int to, nxt;
}e[maxn];
int head[maxn], tot;
int top[maxn];// top[v]表示v所在的重链的顶点
int fa[maxn];
int dep[maxn];
int num[maxn];// num[v]表示以v为根的子树大小
int p[maxn];// p[v]表示v与其父亲节点在线段树的位置
int fp[maxn];
int son[maxn];// 重儿子
int pos;
int n; void init() {
tot = 0; memset(head, -1, sizeof(head));
pos = 1; memset(son, -1, sizeof(son));
}
void addedge(int u, int v) {
e[tot].to = v; e[tot].nxt = head[u]; head[u] = tot++;
} void dfs1(int u, int pre, int d) {
dep[u] = d; fa[u] = pre; num[u] = 1;
for (int i = head[u]; i != -1; i = e[i].nxt) {
int v = e[i].to;
if (v != pre) {
dfs1(v, u, d + 1);
num[u] += num[v];
if (son[u] == -1 || num[v] > num[son[u]])son[u] = v;
}
}
} void dfs2(int u, int sp) {
top[u] = sp;
if (son[u] != -1) {
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 != -1; i = e[i].nxt) {
int v = e[i].to;
if (v != son[u] && v != fa[u])dfs2(v, v);
}
} #define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
int maxx[maxn];
int val[maxn];
void pushup(int rt) {
maxx[rt] = max(maxx[rt << 1], maxx[rt << 1 | 1]);
}
void build(int l, int r, int rt) {
if (l == r) {
maxx[rt] = val[l]; return;
}
int m = (l + r) >> 1;
build(lson); build(rson); pushup(rt);
}
void upd(int p, int x, int l, int r, int rt) {
if (l == r) {
maxx[rt] = x; return;
}
int m = (l + r) >> 1;
if (p <= m)upd(p, x, lson);
else upd(p, x, rson);
pushup(rt);
}
int query(int L, int R, int l, int r, int rt) {
if (L <= l && r <= R) {
return maxx[rt];
}
int m = (l + r) >> 1;
int ans = 0;
if (L <= m)ans = max(ans, query(L, R, lson));
if (m < R)ans = max(ans, query(L, R, rson));
return ans;
} int Find(int u, int v) {
int f1 = top[u], f2 = top[v];
int tmp = 0;
while (f1 != f2) {
if (dep[f1] < dep[f2]) {
swap(f1, f2); swap(u, v);
}
tmp = max(tmp, query(p[f1], p[u], 1, n, 1));
u = fa[f1]; f1 = top[u];
}
if (u == v)return tmp;
if (dep[u] > dep[v])swap(u, v);
return max(tmp, query(p[son[u]], p[v], 1, n, 1));
}
int ed[maxn][3]; int main()
{
// ios::sync_with_stdio(0);
int t; t = rd();
while (t--) {
init(); n = rd();
// getchar();
for (int i = 0; i < n - 1; i++) {
ed[i][0] = rd(); ed[i][1] = rd(); ed[i][2] = rd();
addedge(ed[i][0], ed[i][1]);
addedge(ed[i][1], ed[i][0]);
}
dfs1(1, 0, 0); dfs2(1, 1);
for (int i = 0; i < n - 1; i++) {
if (dep[ed[i][0]] < dep[ed[i][1]]) {
swap(ed[i][0], ed[i][1]);
}
val[p[ed[i][0]]] = ed[i][2];
}
build(1, n, 1);
char opt[10];
while (scanf("%s",opt)) {
if (opt[0] == 'D')break;
int u, v; u = rd(); v = rd();
if (opt[0] == 'Q') {
printf("%d\n", Find(u, v));
}
else upd(p[ed[u - 1][0]], v, 1, n, 1);
}
}
return 0;
}
Query on a tree 树链剖分 [模板]的更多相关文章
- 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 ...
- Hdu 5274 Dylans loves tree (树链剖分模板)
Hdu 5274 Dylans loves tree (树链剖分模板) 题目传送门 #include <queue> #include <cmath> #include < ...
- 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 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 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, ...
- 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 ...
- 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 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 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 ...
随机推荐
- ansible基本使用
ansible介绍 基础概念 ansible是个配置管理工具,可以批量处理一些任务.ansible只需要依赖ssh即可使用,而不需要在受管主机上安装客户端工具. ansible具有幂等性,即以结果为导 ...
- c++多线程编程(二)
这是道面试题目:有三个线程分别打印A.B.C,请用多线程编程实现,在屏幕上循环打印10次ABCABC… 见代码: #include <iostream> #include <Wind ...
- mfs测试
续1 6. 参考文献: 6.1 文献 http://sery.blog.51cto.com/10037/263515 田逸 http://bbs.chinaunix.net/thread-16438 ...
- c++原型模式(Prototype)
原型模式是通过已经存在的对象的接口快速方便的创建新的对象. #include <iostream> #include <string> using namespace std; ...
- cocos2D-x demo 的源码分析 #define ..##.. 的妙用.
最近在看cocos2d-x 但不知道如何下手,于是先看一下他编译的完成的testcpp的源码.发现了下面一段程序 typedef CCLayer* (*NEWTESTFUNC)(); #define ...
- 面试题:JVM类加载机制详解(一)JVM类加载过程 背1
首先Throws(抛出)几个自己学习过程中一直疑惑的问题: 1.什么是类加载?什么时候进行类加载? 2.什么是类初始化?什么时候进行类初始化? 3.什么时候会为变量分配内存? 4.什么时候会为变量赋默 ...
- 一次shell中seq的处理
一次shell中seq的处理 背景:用要shell 提取 文件中内容,文件名是用序列号如下生成,文件差不多有将近400多w个 如下: www.ahlinux.com 原始脚本#! /bin/sh# ...
- SpringMVC——视图和视图解析器
请求处理方法执行完成后,最终返回一个 ModelAndView对象.对于那些返回 String,View 或 ModeMap 等类型的处理方法,Spring MVC 也会在内部将它们装配成一个Mode ...
- 【Azure Active Directory】单一登录 (SAML 协议)
Azure Active Directory 支持 SAML 2.0 Web 浏览器单一登录 (SSO) 配置文件. 若要请求 Azure Active Directory 对用户进行身份验证时,云服 ...
- HDU 3368 Reversi (暴力,DFS)
题意:给定一个8*8的棋盘,然后要懂黑白棋,现在是黑棋走了,问你放一个黑子,最多能翻白子多少个. 析:我是这么想的,反正才是8*8的棋盘,那么就暴吧,反正不会超时,把每一个格能暴力的都暴力,无非是上, ...