Tree Cutting
Tree Cutting
Byteasar has a tree TTT with nnn vertices conveniently labeled with 1,2,...,n1,2,...,n1,2,...,n. Each vertex of the tree has an integer value viv_ivi.
The value of a non-empty tree TTT is equal to v1⊕v2⊕...⊕vnv_1\oplus v_2\oplus ...\oplus v_nv1⊕v2⊕...⊕vn, where ⊕\oplus⊕ denotes bitwise-xor.
Now for every integer kkk from [0,m)[0,m)[0,m), please calculate the number of non-empty subtree of TTT which value are equal to kkk.
A subtree of TTT is a subgraph of TTT that is also a tree.
The first line of the input contains an integer T(1≤T≤10)T(1\leq T\leq10)T(1≤T≤10), denoting the number of test cases.
In each test case, the first line of the input contains two integers n(n≤1000)n(n\leq 1000)n(n≤1000) and m(1≤m≤210)m(1\leq m\leq 2^{10})m(1≤m≤210), denoting the size of the tree TTT and the upper-bound of vvv.
The second line of the input contains nnn integers v1,v2,v3,...,vn(0≤vi<m)v_1,v_2,v_3,...,v_n(0\leq v_i < m)v1,v2,v3,...,vn(0≤vi<m), denoting the value of each node.
Each of the following n−1n-1n−1 lines contains two integers ai,bia_i,b_iai,bi, denoting an edge between vertices aia_iai and bi(1≤ai,bi≤n)b_i(1\leq a_i,b_i\leq n)bi(1≤ai,bi≤n).
It is guaranteed that mmm can be represent as 2k2^k2k, where kkk is a non-negative integer.
For each test case, print a line with mmm integers, the iii-th number denotes the number of non-empty subtree of TTT which value are equal to iii.
The answer is huge, so please module 109+710^9+7109+7.
2
4 4
2 0 1 3
1 2
1 3
1 4
4 4
0 1 3 1
1 2
1 3
1 4
3 3 2 3
2 4 2 3 分析:dp[i][j]表示以i为根异或值为j的方案数;
在加入i的儿子x的子树方案时,dp[i][j]=dp[i][j]+dp[i][k]*dp[x][t](k^t=j);
其中dp[i][k]*dp[x][t](k^t=j)的复杂度为n²,可以用异或卷积加速到nlogn;
代码:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define vi vector<int>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define rev (mod+1)/2
#define inf 0x3f3f3f3f
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
using namespace std;
const int maxn=1e3+;
int n,m,k,t,a[maxn],dp[maxn][maxn],tmp[maxn],ans[maxn];
vi e[maxn];
void fwt(int *a,int n)
{
for(int d=;d<n;d<<=)
for(int m=d<<,i=;i<n;i+=m)
for(int j=;j<d;j++)
{
int x=a[i+j],y=a[i+j+d];
a[i+j]=(x+y)%mod,a[i+j+d]=(x-y+mod)%mod;
}
}
void ufwt(int *a,int n)
{
for(int d=;d<n;d<<=)
for(int m=d<<,i=;i<n;i+=m)
for(int j=;j<d;j++)
{
int x=a[i+j],y=a[i+j+d];
a[i+j]=1LL*(x+y)*rev%mod,a[i+j+d]=(1LL*(x-y)*rev%mod+mod)%mod;
}
}
void solve(int *a,int *b,int n)
{
fwt(a,n);
fwt(b,n);
for(int i=;i<n;i++)a[i]=1LL*a[i]*b[i]%mod;
ufwt(a,n);
}
void dfs(int now,int pre)
{
dp[now][a[now]]=;
for(int x:e[now])
{
if(x==pre)continue;
dfs(x,now);
for(int i=;i<m;i++)
tmp[i]=dp[now][i];
solve(dp[now],dp[x],m);
for(int i=;i<m;i++)
dp[now][i]=(dp[now][i]+tmp[i])%mod;
}
for(int i=;i<m;i++)
ans[i]=(ans[i]+dp[now][i])%mod;
}
int main()
{
int i,j;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
rep(i,,n)
{
scanf("%d",&a[i]),e[i].clear();
rep(j,,m-)dp[i][j]=;
}
rep(i,,m-)ans[i]=;
rep(i,,n-)
{
scanf("%d%d",&j,&k);
e[j].pb(k),e[k].pb(j);
}
dfs(,);
rep(i,,m-)printf("%d%c",ans[i],i<m-?' ':'\n');
}
//system ("pause");
return ;
}
Tree Cutting的更多相关文章
- 【HDU 5909】 Tree Cutting (树形依赖型DP+点分治)
Tree Cutting Problem Description Byteasar has a tree T with n vertices conveniently labeled with 1,2 ...
- BZOJ3391: [Usaco2004 Dec]Tree Cutting网络破坏
3391: [Usaco2004 Dec]Tree Cutting网络破坏 Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 47 Solved: 37[ ...
- BZOJ 3391: [Usaco2004 Dec]Tree Cutting网络破坏( dfs )
因为是棵树 , 所以直接 dfs 就好了... ---------------------------------------------------------------------------- ...
- 3391: [Usaco2004 Dec]Tree Cutting网络破坏
3391: [Usaco2004 Dec]Tree Cutting网络破坏 Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 76 Solved: 59[ ...
- hdu 5909 Tree Cutting [树形DP fwt]
hdu 5909 Tree Cutting 题意:一颗无根树,每个点有权值,连通子树的权值为异或和,求异或和为[0,m)的方案数 \(f[i][j]\)表示子树i中经过i的连通子树异或和为j的方案数 ...
- POJ 2378.Tree Cutting 树形dp 树的重心
Tree Cutting Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 4834 Accepted: 2958 Desc ...
- 【HDU5909】Tree Cutting(FWT)
[HDU5909]Tree Cutting(FWT) 题面 vjudge 题目大意: 给你一棵\(n\)个节点的树,每个节点都有一个小于\(m\)的权值 定义一棵子树的权值为所有节点的异或和,问权值为 ...
- HDU 5909 Tree Cutting 动态规划 快速沃尔什变换
Tree Cutting 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5909 Description Byteasar has a tree T ...
- POJ 2378 Tree Cutting 3140 Contestants Division (简单树形dp)
POJ 2378 Tree Cutting:题意 求删除哪些单点后产生的森林中的每一棵树的大小都小于等于原树大小的一半 #include<cstdio> #include<cstri ...
随机推荐
- jQuery技巧大放送【转】
1.关于页面元素的引用 通过jquery的$()引用元素包括通过id.class.元素名以及元素的层级关系及dom或者xpath条件等方法,且返回的对象为jquery对象(集合对象),不能直接调用do ...
- Memcached 集群的高可用(HA)架构
Memcache自身并没有实现集群功能,如果想用Memcahce实现集群需要借助第三方软件或者自己设计编程实现,这里将采用memagent代理实现,memagent又名magent,大家注意下,不要将 ...
- Ubuntu14.04下安装redis
1.首先在官网上下载redis压缩包 redis-3.2.0.tar.gz 2.解压到到当前文件夹(这里可以解压到随意位置) tar zvxf redis-3.2.0.tar.gz 3.切换到redi ...
- Sersync同步过滤.svn文件夹
Sersync同步过滤.svn文件夹 <filter start="true"> <exclude expression="(.*).svn(.*)&q ...
- 自定义alert和confirm
var common = {}; common.showAlert = function (msg) { var html = "<div id='dialog_alert' clas ...
- php 实现简易模板引擎
1.MVC简介 MVC 是一种使用 MVC(Model View Controller 模型-视图-控制器)设计创建 Web 应用程序的模式(详情自己百度): 1. Model(模型)表示应用程序核心 ...
- 一个初学者的辛酸路程-Python基础-3
前言 不要整天沉迷于学习-. 字典 一.我想跟你聊聊字典 1.为何要有字典? 大家有没有想过为什么要有字典?有列表不就可以了吗? 也许大家会这么认为,我给大家举个例子,大家就明白了. 比如说,我通讯录 ...
- insertAdjacentHTML方法详解
添加HTML内容与文本内容以前用的是innerHTML与innerText方法, 最近发现还有insertAdjacentHTML和 insertAdjacentText方法, 这两个方法更灵活,可以 ...
- grub4dos新手指南-2
Grub4dos 新手指南 一.GRUB4DOS的配置文件Grub4dos 有三个文件,grldr.grldr.mbr和menu.lst,配置文件是menu.lst,和GRUB一样.该文件一般放在和启 ...
- JPA的介绍
一.JPA概述 1.JPA是什么? JPA:Java Persistence API:用于对象持久化的 API,JPA是Java EE 5.0 平台标准的 ORM 规范, 使得应用程序以统一的方式访问 ...