poj 3237 Tree(树链拆分)
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ab路径上节点权值的最大值。
解题思路:树链剖分。然后用线段树维护节点权值,成端更新查询。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 10005;
const int INF = 0x3f3f3f3f;
int id, far[maxn], son[maxn], cnt[maxn], idx[maxn], top[maxn], dep[maxn];
int N, en, first[maxn], jump[maxn * 2], val[maxn];
struct Edge {
int u, v, w;
void set (int u, int v, int w) {
this->u = u;
this->v = v;
this->w = w;
}
}ed[maxn * 2];
void dfs_fir (int u, int pre, int d) {
dep[u] = d;
far[u] = pre;
cnt[u] = 1;
son[u] = 0;
for (int i = first[u]; i + 1; i = jump[i]) {
int v = ed[i].v;
if (v == pre)
continue;
dfs_fir(v, u, d + 1);
cnt[u] += cnt[v];
if (cnt[son[u]] < cnt[v])
son[u] = v;
}
}
void dfs_sec(int u, int rot) {
idx[u] = id++;
top[u] = rot;
if (son[u])
dfs_sec(son[u], rot);
for (int i = first[u]; i + 1; i = jump[i]) {
int v = ed[i].v;
if (v == far[u] || v == son[u])
continue;
dfs_sec(v, v);
}
}
#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int lc[maxn << 2], rc[maxn << 2], filp[maxn << 2], Ma[maxn << 2], Mi[maxn << 2];
inline void maintain (int u) {
filp[u] ^= 1;
swap(Ma[u], Mi[u]);
Ma[u] = -Ma[u];
Mi[u] = -Mi[u];
}
inline void pushup(int u) {
Ma[u] = max(Ma[lson(u)], Ma[rson(u)]);
Mi[u] = min(Mi[lson(u)], Mi[rson(u)]);
}
inline void pushdown (int u) {
if (filp[u]) {
maintain(lson(u));
maintain(rson(u));
filp[u] = 0;
}
}
void build (int u, int l, int r) {
lc[u] = l;
rc[u] = r;
filp[u] = 0;
if (l == r) {
Ma[u] = Mi[u] = val[l];
return;
}
int mid = (l + r) / 2;
build(lson(u), l, mid);
build(rson(u), mid + 1, r);
pushup(u);
}
void modify (int u, int x, int w) {
if (lc[u] == x && rc[u] == x) {
Ma[u] = Mi[u] = w;
return;
}
pushdown(u);
int mid = (lc[u] + rc[u]) / 2;
if (x <= mid)
modify(lson(u), x, w);
else
modify(rson(u), x, w);
pushup(u);
}
void splay(int u, int l, int r) {
if (l <= lc[u] && rc[u] <= r) {
maintain(u);
return;
}
pushdown(u);
int mid = (lc[u] + rc[u]) / 2;
if (l <= mid)
splay(lson(u), l, r);
if (r > mid)
splay(rson(u), l, r);
pushup(u);
}
int query (int u, int l, int r) {
if (l <= lc[u] && rc[u] <= r)
return Ma[u];
pushdown(u);
int mid = (lc[u] + rc[u]) / 2, ret = -INF;
if (l <= mid)
ret = max(ret, query(lson(u), l, r));
if (r > mid)
ret = max(ret, query(rson(u), l, r));
pushup(u);
return ret;
}
inline void add_Edge(int u, int v, int w) {
ed[en].set(u, v, w);
jump[en] = first[u];
first[u] = en++;
}
void init () {
en = 0;
id = 1;
memset(first, -1, sizeof(first));
scanf("%d", &N);
int u, v, w;
for (int i = 0; i < N - 1; i++) {
scanf("%d%d%d", &u, &v, &w);
add_Edge(u, v, w);
add_Edge(v, u, w);
}
dfs_fir(1, 0, 0);
dfs_sec(1, 1);
for (int i = 0; i < N - 1; i++) {
int t = i * 2;
if (dep[ed[t].u] < dep[ed[t].v])
swap(ed[t].u, ed[t].v);
val[idx[ed[t].u]] = ed[t].w;
}
build(1, 1, N);
}
int query (int u, int v) {
int p = top[u], q = top[v], ret = -INF;
while (p != q) {
if (dep[p] < dep[q]) {
swap(p, q);
swap(u, v);
}
ret = max(ret, query(1, idx[p], idx[u]));
u = far[p];
p = top[u];
}
//printf("%d %d\n", u, v);
if (u == v)
return ret;
if (dep[u] > dep[v])
swap(u, v);
ret = max(ret, query(1, idx[son[u]], idx[v]));
return ret;
}
void modify (int u, int v) {
int p = top[u], q = top[v];
while (p != q) {
if (dep[p] < dep[q]) {
swap(p, q);
swap(u, v);
}
splay(1, idx[p], idx[u]);
u = far[p];
p = top[u];
}
if (u == v)
return;
if (dep[u] > dep[v])
swap(u, v);
splay(1, idx[son[u]], idx[v]);
}
void solve () {
int u, v;
char op[20];
while (scanf("%s", op), strcmp(op, "DONE") != 0) {
scanf("%d%d", &u, &v);
if (op[0] == 'C')
modify(1, idx[ed[u*2-2].u], v);
else if (op[0] == 'Q')
printf("%d\n", query(u, v));
else
modify(u, v);
}
}
int main () {
int cas;
scanf("%d", &cas);
while (cas--) {
init();
solve();
}
return 0;
}
版权声明:本文博客原创文章。博客,未经同意,不得转载。
poj 3237 Tree(树链拆分)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 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 ...
- 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 ...
- hdu5044 Tree 树链拆分,点细分,刚,非递归版本
hdu5044 Tree 树链拆分.点细分.刚,非递归版本 //#pragma warning (disable: 4786) //#pragma comment (linker, "/ST ...
- hdu 4912 Paths on the tree(树链拆分+贪婪)
题目链接:hdu 4912 Paths on the tree 题目大意:给定一棵树,和若干个通道.要求尽量选出多的通道,而且两两通道不想交. 解题思路:用树链剖分求LCA,然后依据通道两端节点的LC ...
- poj 3237 Tree [LCA] (树链剖分)
poj 3237 tree inline : 1. inline 定义的类的内联函数,函数的代码被放入符号表中,在使用时直接进行替换,(像宏一样展开),没有了调用的开销,效率也很高. 2. 很明显,类 ...
- 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 < ...
随机推荐
- poj1651(区间dp)
题目连接:http://poj.org/problem?id=1651 题意:给出一组N个数,每次从中抽出一个数(第一和最后一个不能抽),该次的得分即为抽出的数与相邻两个数的乘积.直到只剩下首尾两个数 ...
- tudou link
http://www.tudou.com/programs/view/QdOktCIUfQ0/?tid=-1&aid=-120137222&pid=41050010&oid=2 ...
- ESXI主机打开shell后主机警告处理
昨天为了配置snmp监控,将几台ESXI 5.5主机的shell 在控制台上从disable状态修改为enable状态后,登陆vcenter后,发现所有的主机都有警告. 处理过程如下: 选中有警告标志 ...
- yum 安装软件时报错
报错信息 Another app is currently holding the yum lock; waiting for it to exit 处理方法 rm -rf /var/run/yum. ...
- hdu1166(线段树)
题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=1166 线段树功能:update:单点增减 query:区间求和 #pragma comment(lin ...
- POJ 2942 Knights of the Round Table - from lanshui_Yang
Description Being a knight is a very attractive career: searching for the Holy Grail, saving damsels ...
- AMDU恢复ASM磁盘组数据(測)
--umount ASMCMD> umoung -a asmdg commands: md_backup, md_restor lsattr, setattr ...
- 采用Bash脚本性能监控过程
为一个Linux过程监控,采用Bash脚本. 采用ps命令的过程监控,使用周期加上连续监测的睡眠时间. 使用方法: psmonitor.sh -p [pid] -d [interval] -n [st ...
- java 类 及其 执行过程
java 类 命名 java类文件中 只能有一个公开类 且 公开类类名与当前类文件的文件名一致 方法如果使用了static修饰,那么此方法是类方法,可以 类名.方法名 使用. 你的main方法在同 ...
- Linux虚拟文件系统VFS解决
参考<Linux内核设计与实现> 虚拟文件系统(VFS)它是linux核心和详细I/O一个普通的访问接口之间的包装设备,通过这层界面,linux内核能够以同一的方式訪问各种I/O设备. 虚 ...