POJ 3723 Tree(树链剖分)
POJ 3237 Tree
就多一个取负操作,所以线段树结点就把最大和最小值存下来,每次取负的时候,最大和最小值取负后。交换就可以
代码:
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std; const int N = 10005;
const int INF = 0x3f3f3f3f;
int dep[N], fa[N], son[N], sz[N], top[N], id[N], idx, val[N];
int first[N], next[N * 2], vv[N * 2], en; struct Edge {
int u, v, val;
Edge() {}
Edge(int u, int v, int val) {
this->u = u;
this->v = v;
this->val = val;
}
} e[N]; void init() {
en = 0;
idx = 0;
memset(first, -1, sizeof(first));
} void add_Edge(int u, int v) {
vv[en] = v;
next[en] = first[u];
first[u] = en++;
} void dfs1(int u, int f, int d) {
dep[u] = d;
sz[u] = 1;
fa[u] = f;
son[u] = 0;
for (int i = first[u]; i + 1; i = next[i]) {
int v = vv[i];
if (v == f) continue;
dfs1(v, u, d + 1);
sz[u] += sz[v];
if (sz[son[u]] < sz[v])
son[u] = v;
}
} void dfs2(int u, int tp) {
id[u] = ++idx;
top[u] = tp;
if (son[u]) dfs2(son[u], tp);
for (int i = first[u]; i + 1; i = next[i]) {
int v = vv[i];
if (v == fa[u] || v == son[u]) continue;
dfs2(v, v);
}
} #define lson(x) ((x<<1)+1)
#define rson(x) ((x<<1)+2) struct Node {
int l, r, Max, Min;
bool neg;
} node[N * 4]; void pushup(int x) {
node[x].Max = max(node[lson(x)].Max, node[rson(x)].Max);
node[x].Min = min(node[lson(x)].Min, node[rson(x)].Min);
} void pushdown(int x) {
if (node[x].neg) {
node[lson(x)].Max = -node[lson(x)].Max;
node[rson(x)].Max = -node[rson(x)].Max;
node[lson(x)].Min = -node[lson(x)].Min;
node[rson(x)].Min = -node[rson(x)].Min;
swap(node[lson(x)].Max, node[lson(x)].Min);
swap(node[rson(x)].Max, node[rson(x)].Min);
node[lson(x)].neg = !node[lson(x)].neg;
node[rson(x)].neg = !node[rson(x)].neg;
node[x].neg = false;
}
} void build(int l, int r, int x = 0) {
node[x].l = l; node[x].r = r; node[x].neg = false;
if (l == r) {
node[x].Max = node[x].Min = val[l];
return;
}
int mid = (l + r) / 2;
build(l, mid, lson(x));
build(mid + 1, r, rson(x));
pushup(x);
} void change(int v, int val, int x = 0) {
if (node[x].l == node[x].r) {
node[x].Max = node[x].Min = val;
return;
}
int mid = (node[x].l + node[x].r) / 2;
pushdown(x);
if (v <= mid) change(v, val, lson(x));
if (v > mid) change(v, val, rson(x));
pushup(x);
} void negate(int l, int r, int x = 0) {
if (node[x].l >= l && node[x].r <= r) {
node[x].neg = !node[x].neg;
node[x].Max = -node[x].Max;
node[x].Min = -node[x].Min;
swap(node[x].Max, node[x].Min);
return;
}
int mid = (node[x].l + node[x].r) / 2;
pushdown(x);
if (l <= mid) negate(l, r, lson(x));
if (r > mid) negate(l, r, rson(x));
pushup(x);
} int query(int l, int r, int x = 0) {
if (node[x].l >= l && node[x].r <= r) {
return node[x].Max;
}
int mid = (node[x].l + node[x].r) / 2;
pushdown(x);
int ans = -INF;
if (l <= mid) ans = max(ans, query(l, r, lson(x)));
if (r > mid) ans = max(ans, query(l, r, rson(x)));
pushup(x);
return ans;
} void gao1(int u, int v) {
int tp1 = top[u], tp2 = top[v];
while (tp1 != tp2) {
if (dep[tp1] < dep[tp2]) {
swap(tp1, tp2);
swap(u, v);
}
negate(id[tp1], id[u]);
u = fa[tp1];
tp1 = top[u];
}
if (u == v) return;
if (dep[u] > dep[v]) swap(u, v);
negate(id[son[u]], id[v]);
} int gao2(int u, int v) {
int ans = -INF;
int tp1 = top[u], tp2 = top[v];
while (tp1 != tp2) {
if (dep[tp1] < dep[tp2]) {
swap(tp1, tp2);
swap(u, v);
}
ans = max(ans, query(id[tp1], id[u]));
u = fa[tp1];
tp1 = top[u];
}
if (u == v) return ans;
if (dep[u] > dep[v]) swap(u, v);
ans = max(ans, query(id[son[u]], id[v]));
return ans;
} int T, n; int main() {
scanf("%d", &T);
while (T--) {
init();
scanf("%d", &n);
int u, v, w;
for (int i = 1; i < n; i++) {
scanf("%d%d%d", &u, &v, &w);
add_Edge(u, v);
add_Edge(v, u);
e[i] = Edge(u, v, w);
}
dfs1(1, 0, 1);
dfs2(1, 1);
for (int i = 1; i < n; i++) {
if (dep[e[i].u] < dep[e[i].v]) swap(e[i].u, e[i].v);
val[id[e[i].u]] = e[i].val;
}
build(1, idx);
char Q[10];
int a, b;
while (scanf("%s", Q)) {
if (Q[0] == 'D') break;
scanf("%d%d", &a, &b);
if (Q[0] == 'Q') printf("%d\n", gao2(a, b));
if (Q[0] == 'C') change(id[e[a].u], b);
if (Q[0] == 'N') gao1(a, b);
}
}
return 0;
}
POJ 3723 Tree(树链剖分)的更多相关文章
- poj 3237 Tree 树链剖分
题目链接:http://poj.org/problem?id=3237 You are given a tree with N nodes. The tree’s nodes are numbered ...
- POJ 3237.Tree -树链剖分(边权)(边值更新、路径边权最值、区间标记)贴个板子备忘
Tree Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 12247 Accepted: 3151 Descriptio ...
- poj 3237 Tree 树链剖分+线段树
Description You are given a tree with N nodes. The tree’s nodes are numbered 1 through N and its edg ...
- POJ 3237 Tree (树链剖分 路径剖分 线段树的lazy标记)
题目链接:http://poj.org/problem?id=3237 一棵有边权的树,有3种操作. 树链剖分+线段树lazy标记.lazy为0表示没更新区间或者区间更新了2的倍数次,1表示为更新,每 ...
- POJ3237 Tree 树链剖分 边权
POJ3237 Tree 树链剖分 边权 传送门:http://poj.org/problem?id=3237 题意: n个点的,n-1条边 修改单边边权 将a->b的边权取反 查询a-> ...
- 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 < ...
- poj 3237 Tree(树链拆分)
题目链接:poj 3237 Tree 题目大意:给定一棵树,三种操作: CHANGE i v:将i节点权值变为v NEGATE a b:将ab路径上全部节点的权值变为相反数 QUERY a b:查询a ...
- 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-4353】Play with tree 树链剖分
4353: Play with tree Time Limit: 20 Sec Memory Limit: 256 MBSubmit: 31 Solved: 19[Submit][Status][ ...
- 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, ...
随机推荐
- getField();在TP5里成什么了?
拆分为value和column了 $comps=db("company")->where(array("areaid"=>$areaid))-> ...
- Win10,JDK8,tomact7.0.85配置
今天在win10上配置了jdk以及tomact先前使用jdk7+tomact7.0.56运行失败. 后经调整后正确配置注意事项如下: 1.下载并按照jdk-8u161-windows-x64,默认安装 ...
- Java 基本的递归写法
1.首先我们得有一个树状结构的表,类似这种结构.必须得有 id,pid 其他的根据需要来. 我们叫它treeTbl表吧.这里pid为0的表示是顶级节点. 2.接着select * from tree ...
- CSS浮动的处理
之前已经发过一遍有关浮动的解决办法,今天看到一些资料后又有了新的想法: 在CSS布局中float属性经常会被用到,但使用float属性后会使其在普通流中脱离父容器,让人很苦恼 1 浮动带来布局的便利, ...
- CentOS 安装dotNetCore
如果要在CentOS上运行.net Core程序,必须安装.net Core Sdk 具体安装 方法,可以参考微软官方站点说明,非常详细: 1)百度搜索 .Net Core 2)先择CentOS版本: ...
- Oracle SQL 性能优化技巧
Select语句完整的执行顺序: SQL Select语句完整的执行顺序: 1. from子句组装来自不同数据源的数据: 2.where子句基于指定的条件对记录行进行筛选: 3.group by子句将 ...
- 国内DNS服务器地址
114DNS114.114.114.114114.114.115.115 腾讯119.29.29.29 百度180.76.76.76 阿里223.5.5.5223.6.6.6 [THE END]
- selenium菜单操作
连接到前端这个菜单下面的HTML/CSS子菜单 driver.get("https://www.imooc.com"); WebElement login = driver.fin ...
- 12 Python+selenium对日期控件进行处理(采用执行JS脚本)
[环境信息] Python34+IE+windows2008 [说明] 1.对于日期控件,没有办法通过定位元素再直接传值的方式处理.可以采用执行JavaScript处理. PS:还要去学学js怎么写, ...
- MVC控制器返回值
public ActionResult Index(string id)//主页 //参数string searchString 访问方式为index?searchString=xxxx .参数str ...