• ## Trees

2020-12-04 22:31:01
Do you think this could be applied to trees as well? Imagine I have a tree of lightweight structs. Can I compare two trees and find the minimum changes required going from one to the other?</p><p>该...
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• [索引页] Expression trees语言级别的代码呈现数据形式....下面图显示是一个Expression的事例和它呈现的一个Expression trees的形式.Expression不同部分是匹配相应的Expression trees子节点的有颜色的代...

[索引页]
Expression trees语言级别的代码呈现数据形式.这个数据被存储在一个数形结构中.在Expression trees中每一个节点呈现一个表达式,如一个方法调用或二元运算符.
下面图显示是一个Expression的事例和它呈现的一个Expression trees的形式.Expression不同部分是匹配相应的Expression trees子节点的有颜色的代码.
还有Expression trees子节点的类型不同.
下面的事例代码示范怎样表达树呈现lambda表达式 num => num < 5 (C#) or Function(num) num < 5 (Visual Basic)被分解为几部分.// Add the following using directive to your code file:// using System.Linq.Expressions;// Create an expression tree.Expression<Func<int, bool>> exprTree = num => num < 5;// Decompose the expression tree.ParameterExpression param = (ParameterExpression)exprTree.Parameters[0];BinaryExpression operation = (BinaryExpression)exprTree.Body;ParameterExpression left = (ParameterExpression)operation.Left;ConstantExpression right = (ConstantExpression)operation.Right;Console.WriteLine("Decomposed expression: {0} => {1} {2} {3}",                  param.Name, left.Name, operation.NodeType, right.Value);/**//*  This code produces the following output:    Decomposed expression: num => num LessThan 5*/

Building Expression Trees
在System.Linq.Expressions 命名空间里提供一个API来手动构建expression trees.The Expression类包括创建表达树指定子节点的类型的static factory methods.如一个ParameterExpression,它是呈现一个表达式参数名,或一个MethodCallExpression,而这一个是呈现一个调用方法.ParameterExpression, MethodCallExpression,和其他的详细的指定表达树类型的expression定义在System.Linq.Expressions命名空间中.这些类型都派生于抽象类型Expression.
编译器还能自动为你构建一个expression tree.一个编译器产生表达式树是始终来源于一个子节点的Expression(TDelegate)类型;就是说,它的根节点的子节点呈现lambda表达式.
下面代码事例是以两个方式创建一个呈现ambda expression num => num < 5 (C#) or Function(num) num < 5 (Visual Basic)的表达式树.// Add the following using directive to your code file:// using System.Linq.Expressions;// Manually build the expression tree for// the lambda expression num => num < 5.ParameterExpression numParam = Expression.Parameter(typeof(int), "num");ConstantExpression five = Expression.Constant(5, typeof(int));BinaryExpression numLessThanFive = Expression.LessThan(numParam, five);Expression<Func<int, bool>> lambda1 =    Expression.Lambda<Func<int, bool>>(        numLessThanFive,        new ParameterExpression[] { numParam });// Let the compiler generate the expression tree for// the lambda expression num => num < 5.Expression<Func<int, bool>> lambda2 = num => num < 5;

Immutability of Expression Trees
Expression trees是不可变化的.这个意思是如果你要修改一个表达树,你必须copey和修改现有的表达式树构建一个新的表达式树.你能使用一个表达式树访问者来跨越到其他表达式树上.
Lambda Expressions
当一个lambda expression 被分配一个Expression(TDelegate)类型的变量,这个编译器就会发布一个呈现lambda表达式的表达式树.例如一些标准查询操作符(standard query operator methods)方法被定义在Queryable类中且已有的参数类型为Expression(TDelegate).你能通过在lambda expression中调用这些方法并且编译器将产生一个表达式树.
而这里的Expression(TDelegate)类型提供一个Compile方法,这个方法是将呈现表达式树的代码编译成为一个可执行的委托.这个可执行的代码是相当于将代码按照已有lambda expression分配的最初委托类型来生成.
注意:
仅仅这些expression trees呈现的功能,也就是Expression(TDelegate)和它父类型LambdaExpression,能被编译成执行代码,你必须首先在Lambda Expression 的子节点约束他们执行在其他的表达树的类型.你能按照调用Lambda方法和通过表达式树的参数来获得这样的Lambda Expression. 原文地址
-------------worksguo
转载于:https://www.cnblogs.com/worksguo/archive/2008/03/28/1127069.html
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• <div><p>Before lowering JProfling value calls, scan the trees around to make sure it won't orphan a compressedrefs anchor node. <p>Signed-off-by: Rahil Shah </p><p>该提问来源于开源项目：...
• CodeForces – 711C ... 题目描述 ZS the Coder and Chris the Baboon has arrived at Udayland! They walked in the ... They decided to be naughty and color the trees in the park. The trees are numbered wit..
CodeForces – 711C

https://www.luogu.com.cn/problem/CF711C

题目描述

ZS the Coder and Chris the Baboon has arrived at Udayland! They walked in the park where nn trees grow. They decided to be naughty and color the trees in the park. The trees are numbered with integers from 11 to nn from left to right.

Initially, tree ii has color c_{i}ci​ . ZS the Coder and Chris the Baboon recognizes only mm different colors, so 0<=c_{i}<=m0<=ci​<=m , where c_{i}=0ci​=0 means that tree ii is uncolored.

ZS the Coder and Chris the Baboon decides to color only the uncolored trees, i.e. the trees with c_{i}=0ci​=0 . They can color each of them them in any of the mm colors from 11 to mm . Coloring the ii -th tree with color jj requires exactly p_{i,j}pi,j​ litres of paint.

The two friends define the beauty of a coloring of the trees as the minimum number of contiguous groups (each group contains some subsegment of trees) you can split all the nn trees into so that each group contains trees of the same color. For example, if the colors of the trees from left to right are 2,1,1,1,3,2,2,3,1,32,1,1,1,3,2,2,3,1,3 , the beauty of the coloring is 77 , since we can partition the trees into 77 contiguous groups of the same color : {2},{1,1,1},{3},{2,2},{3},{1},{3}2,1,1,1,3,2,2,3,1,3 .

ZS the Coder and Chris the Baboon wants to color all uncolored trees so that the beauty of the coloring is exactly kk . They need your help to determine the minimum amount of paint (in litres) needed to finish the job.

Please note that the friends can’t color the trees that are already colored.

输入格式

The first line contains three integers, nn , mm and kk ( 1<=k<=n<=1001<=k<=n<=100 , 1<=m<=1001<=m<=100 ) — the number of trees, number of colors and beauty of the resulting coloring respectively.

The second line contains nn integers c_{1},c_{2},…,c_{n}c1​,c2​,…,cn​ ( 0<=c_{i}<=m0<=ci​<=m ), the initial colors of the trees. c_{i}ci​ equals to 00 if the tree number ii is uncolored, otherwise the ii -th tree has color c_{i}ci​ .

Then nn lines follow. Each of them contains mm integers. The jj -th number on the ii -th of them line denotes p_{i,j}pi,j​ ( 1<=p_{i,j}<=10^{9}1<=pi,j​<=109 ) — the amount of litres the friends need to color ii -th tree with color jj . p_{i,j}pi,j​ ‘s are specified even for the initially colored trees, but such trees still can’t be colored.

输出格式

Print a single integer, the minimum amount of paint needed to color the trees. If there are no valid tree colorings of beauty kk , print -1−1 .

题意翻译

题意：

有 nn 棵树， mm 种颜料，要求现在要给这些树涂上颜料，最后涂成 kk 段（连续颜色相同划为一段如 22 , 11 , 11 , 11 , 33 , 22 , 22 , 33 , 11 , 33 是77段），有些树已经涂了，则不涂了只能涂一次，输入nn个数（每个数为00~mm），0表示还没有涂，11~mm表示已经涂了哪种颜料。接下来输入 nn 行 mm 列，表示每棵树涂成每种颜色所要的颜料量。现在要把所有树都涂上颜料涂成 kk 段，求最少要用的颜料量

翻译 byby @Happynewyear

输入输出样例

输入 #1复制

3 2 2
0 0 0
1 2
3 4
5 6

输出 #1复制

10

输入 #2复制

3 2 2
2 1 2
1 3
2 4
3 5

输出 #2复制

-1

输入 #3复制

3 2 2
2 0 0
1 3
2 4
3 5

输出 #3复制

5

输入 #4复制

3 2 3
2 1 2
1 3
2 4
3 5

输出 #4复制

0

说明/提示

In the first sample case, coloring the trees with colors 2,1,12,1,1 minimizes the amount of paint used, which equals to 2+3+5=102+3+5=10 . Note that 1,1,11,1,1 would not be valid because the beauty of such coloring equals to 11 ( {1,1,1}1,1,1 is a way to group the trees into a single group of the same color).

In the second sample case, all the trees are colored, but the beauty of the coloring is 33 , so there is no valid coloring, and the answer is -1−1 .

In the last sample case, all the trees are colored and the beauty of the coloring matches kk , so no paint is used and the answer is 00 .

思路：一道线性dp的好题目。

定义：dp[i][j][k]为前i个点分成j段并且最后一段颜色是k的最小代价

当第i个点是有颜色的。

如果和前面一个点颜色一致,dp[i][j][l]=min(dp[i][j][l],dp[i-1][j][l]);

如果和前面一个点的颜色不一致,dp[i][j][col[i]]=min(dp[i][j][col[i]],dp[i-1][j-1][l]);

如果第i个点是无颜色的

如果涂上和前面一个点不一致的颜色。dp[i][j][l]=min(dp[i][j][l],d[i-1][j-1][p]+w[i][l]);

如果涂上和前面一个点一致的颜色。dp[i][j][l]=min(dp[i][j][l],dp[i-1][j][p]+w[i][l]);

注意dp的初始化开大一点。

#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<cstdio>
#include<algorithm>
#define inf 0x7f7f7f7f7f7f
using namespace std;
typedef long long LL;
const int maxn=200;
LL dp[maxn][maxn][maxn];//dp[i][j][k]：前i棵树分成j段且最后一段颜色为k时的最小花费
LL w[maxn][maxn];
LL col[maxn];
LL n,m,k;
int main(void)
{
cin.tie(0);std::ios::sync_with_stdio(false);
cin>>n>>m>>k;
for(LL i=1;i<=n;i++) cin>>col[i];
for(LL i=1;i<=n;i++){
for(LL j=1;j<=m;j++)
cin>>w[i][j];
}
memset(dp,inf,sizeof(dp));
if(col[1]) dp[1][1][col[1]]=0;
else{
for(LL i=1;i<=m;i++) dp[1][1][i]=w[1][i];
}
for(LL i=2;i<=n;i++){
for(LL j=1;j<=i&&j<=k;j++){
if(col[i]){//第i个点有颜色
for(LL l=1;l<=m;l++){
if(col[i]==l){
dp[i][j][l]=min(dp[i][j][l],dp[i-1][j][l]);//同样颜色为连续一段
}
else dp[i][j][col[i]]=min(dp[i][j][col[i]],dp[i-1][j-1][l]);//不同颜色为新开一段
}
}
else{
for(LL l=1;l<=m;l++){
for(LL p=1;p<=m;p++){
if(l==p) dp[i][j][l]=min(dp[i][j][l],dp[i-1][j][p]+w[i][l]);//一样颜色就连续一段
else dp[i][j][l]=min(dp[i][j][l],dp[i-1][j-1][p]+w[i][l]);//不一样颜色就另开一段
}
}
}
}
}
LL ans=inf;
if(col[n]) ans=dp[n][k][col[n]];
else {
for(LL l=1;l<=m;l++){
ans=min(ans,dp[n][k][l]);
}
}
if(!(ans<inf)) cout<<"-1"<<endl;
else cout<<ans<<endl;
return 0;
}


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• There are many trees forming a m * n grid, the grid starts from (1,1). Farmer Sherlock is standing at (0,0) point. He wonders how many trees he can see. If two trees and Sherlock are in one line, ...
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Convertion from Trees, Binary Trees, and Forest树转换为二叉树：森林转化为二叉树二叉树转化为森林
树转换为二叉树：
转换原则
第一个右兄弟结点->右孩子
第一个孩子结点->左孩子

特点
树的前序遍历<—>二叉树的前序遍历
树的后序遍历<—>二叉树的中序遍历
森林转化为二叉树
转换原则
（1）将森林中的每棵树转换成二叉树
（2）从第二棵树开始，依次把后一棵二叉树的根结点作为前一棵二叉树根结点的右孩子。
二叉树转化为森林
转换原则
1）根结点的右孩子  第二棵树的根结点
根结点右孩子的右孩子第三棵树的根结点
……
2）根结点的左孩子
按照结点的左孩子是该结点的第一个孩子，结点右孩子是该结点的右兄弟转换。


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Extremely Randomized Trees(ExrRa Trees)

今天打kaggle的泰坦尼克的时候，看到了这样一个算法：
ExtRa Trees是Extremely Randomized Trees的缩写，意思就是极度随机树。这是一种组合方法，与其说像决策树，实际上它更像随机森林。

与随机森林的相同点 ：1. bootstrap了样本。2. 随机选取了部分特征，来构造一个棵树。

我们知道随机森林是由很多个随机的决策树组成的，extra trees比随机森林更随机。

与随机森林的不同点：每棵决策树选择划分点的方式不同。对于普通决策树，每个特征都是根据某个标准(信息增益或者gini不纯)去进行划分，比如说一个数值特征的取值范围是0到100，当我们把这个特征划分为0-45，45-100的时候，效果最佳（从信息增益或者gini不纯的角度），决策树就会将45作为划分点；然后再按照某个标准来选择特征。而对于extra trees中的决策树，划分点的选择更为随机，比如一个数值特征的取值范围是0到100，我们随机选择一个0到100的数值作为划分点；然后再按照评判标准选择一个特征。
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• <div><p>Hello, Thanks for this wonderful library...<p>I would like the library to support github trees, is it foreseen (https://developer.github.com/v3/git/trees/)? For instance I need to retrieve this: ...
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...