HDU 5274(LCA + 线段树)
Dylans loves tree
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 747 Accepted Submission(s): 144
All nodes have a value A[i].Nodes
on tree is numbered by 1∼N.
Then he is given Q questions
like that:
①0 x y:change
node x′s value
to y
②1 x y:For
all the value in the path from x to y,do
they all appear even times?
For each ② question,it guarantees that there is at most one value that appears odd times on the path.
1≤N,Q≤100000,
the value A[i]∈N and A[i]≤100000
(T≤3 and
there is at most one testcase that N>1000)
For each testcase:
In the first line there are two numbers N and Q.
Then in the next N−1 lines
there are pairs of (X,Y) that
stand for a road from x to y.
Then in the next line there are N numbers A1..AN stand
for value.
In the next Q lines
there are three numbers(opt,x,y).
1
3 2
1 2
2 3
1 1 1
1 1 2
1 1 3
-1
1HintIf you want to hack someone,N and Q in your testdata must smaller than 10000,and you shouldn't print any space in each end of the line.
先考虑无改动的情况,令Xor[i]表示i到根节点路径上的异或和。则随意节点的(u,v)的异或和能够转化为Xor[u]^Xor[v]^a[LCA(u,v)].考虑改动的情况。改动节点u,仅仅会以u为根的子树的Xor值产生影响,由于一颗子树的dfs序是连续的我们非常自然的想到用线段树去维护他,pSeg[u]表示u在dfs序中的位置,siz[u]表示以u为根的子树大小,则这课颗子树相应的区间就是[pSeg[u],pSeg[u]+siz[u]-1],改动的时候仅仅须要将这段区间先异或上原来的值a[u],在异或上要变成的值y,然后改动a[u]
= y;两次异或能够一步到位。直接异或上a[u]^y即可。
#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;
const ll mod = 1e9 + 7;
const int maxn = 1e5 + 10;
#define to first
#define next second
#define foreach(it,v) for(__typeof(v.begin()) it = v.begin(); it != v.end(); ++it)
int pos[maxn],d[20][maxn<<1],wid[maxn<<1],head[maxn];
int a[maxn],depth[maxn],sid,pSeg[maxn],siz[maxn],Xor[maxn];
typedef pair<int,int> Edge;
Edge edges[maxn<<1];
int tot = 0,e = 0;
void AddEdge(int u,int v)
{
edges[++e] = make_pair(v,head[u]);head[u] = e;
edges[++e] = make_pair(u,head[v]);head[v] = e;
}
void pre(int u,int fa,int dep = 0,int Xo = 0)
{
Xo ^= a[u];
Xor[++sid] = Xo;
pSeg[u] = sid;
siz[u] = 1;
d[0][++tot] = u;
if(!pos[u]) {
pos[u] = tot;
depth[u] = dep;
}
for(int i = head[u]; i ; i = edges[i].next) {
int v = edges[i].to;
if(v == fa) continue;
pre(v,u,dep+1,Xo);
siz[u] += siz[v];
d[0][++tot] = u;
}
}
void RMQ_init(int n)
{
for(int i = 1,w = 1; i <= n; i++) {
if((1<<w)<=i) w++;
wid[i] = w - 1;
}
for(int i = 1; (1<<i) <= n; i++) {
for(int j = 1; j + (1<<i) - 1 <= n; j++) {
d[i][j] = depth[d[i-1][j]] < depth[d[i-1][j+(1<<(i-1))]] ? d[i-1][j] : d[i-1][j+(1<<(i-1))];
}
}
}
int LCA(int u,int v)
{
u = pos[u];
v = pos[v];
if(u > v) swap(u,v);
int k = wid[v-u+1];
return depth[d[k][u]] < depth[d[k][v-(1<<k)+1]] ? d[k][u] : d[k][v-(1<<k)+1];
}
int seg[maxn<<2];
int ql,qr,x;
void push_down(int o)
{
seg[o<<1] ^= seg[o];
seg[o<<1|1] ^= seg[o];
seg[o] = 0;
}
void Modify(int o,int L,int R)
{
if(ql<=L&&qr>=R) {
seg[o] ^= x;
return ;
}
push_down(o);
int mid = (L+R)>>1;
if(ql<=mid) Modify(o<<1,L,mid);
if(qr>mid) Modify(o<<1|1,mid+1,R);
}
int Query(int o,int L,int R)
{
if(L == R) {
return Xor[L] ^ seg[o];
}
int mid = (L+R) >>1;
push_down(o);
if(x<=mid)return Query(o<<1,L,mid);
return Query(o<<1|1,mid+1,R);
}
int main(int argc, char const *argv[])
{
int T;scanf("%d",&T);
while(T--) {
int N,Q;scanf("%d%d",&N,&Q);
e = sid = tot = 0;
memset(head,0,sizeof(head[0])*(N+1));
for(int i = 1; i < N; i++) {
int u,v;scanf("%d%d",&u,&v);
AddEdge(u,v);
}
for(int i = 1; i <= N; i++) {
scanf("%d",a+i);
++a[i];
}
pre(1,-1);
RMQ_init(tot);
memset(seg,0,sizeof(seg[0])*(2*N+10));
while(Q--) {
scanf("%d%d%d",&x,&ql,&qr);
if(x==0) {
qr++;
int L = pSeg[ql], R = pSeg[ql] + siz[ql] - 1;
x = a[ql] ^ qr;
a[ql] = qr;
ql = L,qr = R;
Modify(1,1,N);
}else {
x = pSeg[ql];
int ans = Query(1,1,N);
x = pSeg[qr];
ans ^= Query(1,1,N);
ans ^= a[LCA(ql,qr)];
if(ans==0)puts("-1");
else printf("%d\n", ans-1);
}
}
}
return 0;
}
HDU 5274(LCA + 线段树)的更多相关文章
- hdu 5274 Dylans loves tree(LCA + 线段树)
Dylans loves tree Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Othe ...
- hdu 5266 pog loves szh III(lca + 线段树)
I - pog loves szh III Time Limit:6000MS Memory Limit:131072KB 64bit IO Format:%I64d & %I ...
- hdu 4031 attack 线段树区间更新
Attack Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)Total Subm ...
- hdu 4288 离线线段树+间隔求和
Coder Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Su ...
- hdu 3016 dp+线段树
Man Down Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total S ...
- BZOJ 2243: [SDOI2011]染色 树链剖分 倍增lca 线段树
2243: [SDOI2011]染色 Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://www.lydsy.com/JudgeOnline/pr ...
- [51nod 1766]树上的最远点对 (树的直径+ST表求lca+线段树)
[51nod 1766]树上的最远点对 (树的直径+ST表求lca+线段树) 题面 给出一棵N个点的树,Q次询问一点编号在区间[l1,r1]内,另一点编号在区间[l2,r2]内的所有点对距离最大值.\ ...
- 2018多校第十场 HDU 6430 (线段树合并)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6430 题意:一棵树上每个节点权值为v[i],每个节点的heard值是:以它为LCA的两个节点的GCD的 ...
- HDU 5877 dfs+ 线段树(或+树状树组)
1.HDU 5877 Weak Pair 2.总结:有多种做法,这里写了dfs+线段树(或+树状树组),还可用主席树或平衡树,但还不会这两个 3.思路:利用dfs遍历子节点,同时对于每个子节点au, ...
随机推荐
- 使用PoolingHttpClientConnectionManager解决httpclient的多线程请求问题
直接上代码 1.主程序 public class TestMain { public static void main(String[] args) throws NSQException, Time ...
- C#窗体间的跳转传值
1.开发平台VS2012 2.需求:从一个窗体跳转到另一个窗体,并传递参数,接收返回值. 3.案列如图: 4.代码如下: 登陆窗体: //当点击注册按钮 private void button2_Cl ...
- Java基础学习总结(56)——学java必知十大学习目标
诞生至今已有20年的Java,现在依然有很多人使用.回顾过去十五年的成果,Java一直是数一数二的.Java已经成为世界范围内应用最为广泛的编程语言之一.那么在学java的时候你知道到底要学什么吗?一 ...
- ASP.NET-POSTBACK是什么
当我们直接从服务端读取网页时,表时此网页并没有post(提交),当用户再次提交表单时,就会把此网页的相关参数传给服务器处理,对于服务器来说就是一个postback(提交返回),即提交回来了.这就是po ...
- POJ——T1789 Truck History
http://poj.org/problem?id=1789 Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 27597 ...
- spring mvc常用知识点总结
1.spring mvc是靠spring 启动的.通过springjar包的org.springframework.web.servlet.DispatcherServlet这个servlet类具体启 ...
- Python中常见的文件对象内建函数
文件对象内建方法列表 文件对象的方法 操作 file.close() 关闭文件 file.fileno() 返回文件的描写叙述符(file descriptor.FD,整数值) file.flush( ...
- Android开发之AudioManager(音频管理器)具体解释
AudioManager简单介绍: AudioManager类提供了訪问音量和振铃器mode控制. 使用Context.getSystemService(Context.AUDIO_SERVICE)来 ...
- jQuery操作元素的属性与样式
本文学习如何使用jQuery获取和操作元素的属性和CSS样式. 元素属性和Dom属性 对于下面这样一个标签元素: <img id='img' src="1.jpg" alt= ...
- Linux常用命令之rpm安装命令
转自:http://www.cnblogs.com/datasyman/p/6942557.html 在 Linux 操作系统下,几乎所有的软件均通过RPM 进行安装.卸载及管理等操作.RPM 的全称 ...