1053 Path of Equal Weight(30 分) Given a non-empty tree with root R, and with weight W​i​​ assigned to each tree node T​i​​. The weight of a path from R to L is defined to be the sum of the weights of all the nodes along the path from R to any leaf no…

1053 Path of Equal Weight(30 分) Given a non-empty tree with root R, and with weight W​i​​ assigned to each tree node T​i​​. The weight of a path from R to L is defined to be the sum of the weights of all the nodes along the path from R to any leaf no…

1053. Path of Equal Weight (30) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Given a non-empty tree with root R, and with weight Wi assigned to each tree node Ti. The weight of a path from R to L is defined to be the sum of the…

1053 Path of Equal Weight (30 分)   Given a non-empty tree with root R, and with weight W​i​​ assigned to each tree node T​i​​. The weight of a path from R to L is defined to be the sum of the weights of all the nodes along the path from R to any leaf…

1053 Path of Equal Weight 给定一个非空的树,树根为 RR. 树中每个节点 TiTi 的权重为 WiWi. 从 RR 到 LL 的路径权重定义为从根节点 RR 到任何叶节点 LL 的路径中包含的所有节点的权重之和. 现在给定一个加权树以及一个给定权重数字,请你找出树中所有的权重等于该数字的路径(必须从根节点到叶节点). 例如,我们考虑下图的树,对于每个节点,上方的数字是节点 ID,它是两位数字,而下方的数字是该节点的权重. 假设给定数为 2424,则存在 44 个具有相同…

Given a non-empty tree with root R, and with weight Wi assigned to each tree node Ti. The weight of a path from R to Lis defined to be the sum of the weights of all the nodes along the path from R to any leaf node L. Now given any weighted tree, you…

Given a non-empty tree with root R, and with weight W​i​​ assigned to each tree node T​i​​. The weight of a path from Rto L is defined to be the sum of the weights of all the nodes along the path from R to any leaf node L. Now given any weighted tree…

#include <cstdio> #include <cstdlib> #include <vector> #include <algorithm> using namespace std; vector<vector<int>* > paths; class Node { public: vector<int> child; int weight; Node() : weight(w){} }; int str2num…

Given a non-empty tree with root R, and with weight W~i~ assigned to each tree node T~i~. The weight of a path from R to L is defined to be the sum of the weights of all the nodes along the path from R to any leaf node L. Now given any weighted tree,…

题目 Given a non-empty tree with root R, and with weight Wi assigned to each tree node Ti. The weight of a path from R to L is defined to be the sum of the weights of all the nodes along the path from R to any leaf node L. Now given any weighted tree,…

Given a non-empty tree with root R, and with weight W​i​​ assigned to each tree node T​i​​. The weight of a path from R to L is defined to be the sum of the weights of all the nodes along the path from R to any leaf node L. Now given any weighted tre…

Given a non-empty tree with root R, and with weight W​i​​ assigned to each tree node T​i​​. The weight of a path from R to L is defined to be the sum of the weights of all the nodes along the path from R to any leaf node L. Now given any weighted tre…

简单DFS #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<queue> #include<stack> #include<string> #include<algorithm> using namespace std; +; vector<int>Tre…

由于最后输出的路径排序是降序输出,相当于dfs的时候应该先遍历w最大的子节点. 链式前向星的遍历是从最后add的子节点开始,最后添加的应该是w最大的子节点, 因此建树的时候先对child按w从小到大排序,然后再add建边. 水题一个,不多说了. #include <iostream> #include <algorithm> #include <cstdio> #include <string.h> using namespace std; ; int he…

题目大意:给出树的结构和权值,找从根结点到叶子结点的路径上的权值相加之和等于给定目标数的路径,并且从大到小输出路径 #include<bits/stdc++.h> using namespace std; int n,m,sum; ; struct node { int w; vector<int>p; }tree[N]; bool cmp(int a,int b) { return tree[a].w>tree[b].w; } vector<int>path;…

题意: 输入三个正整数N,M,S(N<=100,M<N,S<=2^30)分别代表数的结点个数,非叶子结点个数和需要查询的值,接下来输入N个正整数(<1000)代表每个结点的权重,接下来输入M行,每行包括一个两位数字组成的数代表非叶子结点的编号以及数字x表示它的孩子结点个数,接着输入x个数字表示孩子结点的编号.以非递增序输出从根到叶子结点的路径权重,它们的和等于S. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<…

1053. Path of Equal Weight (30) 时间限制 10 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Given a non-empty tree with root R, and with weight Wi assigned to each tree node Ti. The weight of a path from R to L is defined to be the sum of the…

Path of Equal Weight (DFS)   Given a non-empty tree with root R, and with weight Wi assigned to each tree node Ti. The weight of a path from R to L is defined to be the sum of the weights of all the nodes along the path from R to any leaf node L. Now…

Source: PAT A1053 Path of Equal Weight (30 分) Description: Given a non-empty tree with root R, and with weight W​i​​ assigned to each tree node T​i​​. The weight of a path from R to L is defined to be the sum of the weights of all the nodes along the…

Given a non-empty tree with root R, and with weight Wi assigned to each tree node Ti. The weight of a path from R to L is defined to be the sum of the weights of all the nodes along the path from R to any leaf node L. Now given any weighted tree, you…

Given a non-empty tree with root R, and with weight W​i​​ assigned to each tree node T​i​​. The weight of a path from R to L is defined to be the sum of the weights of all the nodes along the path from R to any leaf node L. Now given any weighted tre…

//之前一直尝试用vector存储path,但是每次错误后回退上一级节点时不能争取回退,导致探索路径正确,但是输出不正确,用参数num,标记前一个路径点的位置传递参数,就好多了 //其中在输入时就将后继节点按照权值大小排列,是可以学习方法,再输入时就处理好,方便dfs探索,有几道PAT题目都是在该基础上对路径加要求,都可以直接改进输出那一部分代码,更新更和要求的路径,全部遍历完,就可得到所求符合要求的的路径 //参数传递的思想也不错,之前不习惯,总是直接把ans存进vector,最后输出 #in…

1001. A+B Format (20) 注意负数,没别的了. 用scanf来补 前导0 和 前导的空格 很方便. #include <iostream> #include <cstdio> using namespace std; ]; int main() { int A,B; cin>>A>>B; A+=B; ) { A=-A; cout<<"-"; } ; while(A) { a[n++]=A%; A/=; } ;…

准备每天刷两题PAT真题.(一句话题解) 1001 A+B Format  模拟输出,注意格式 #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; string ans = ""; int main() { ; cin >> a >> b; c = a + b; ) {…

博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6102219.html特别不喜欢那些随便转载别人的原创文章又不给出链接的所以不准偷偷复制博主的博客噢~~ 时隔两年,又开始刷题啦,这篇用于PAT甲级题解,会随着不断刷题持续更新中,至于更新速度呢,嘿嘿,无法估计,不知道什么时候刷完这100多道题. 带*的是我认为比较不错的题目,其它的难点也顶多是细节处理的问题~ 做着做着,发现有些题目真的是太水了,都不想写题解了…

树(23) 备注 1004 Counting Leaves   1020 Tree Traversals   1043 Is It a Binary Search Tree 判断BST,BST的性质 1053 Path of Equal Weight   1064 Complete Binary Search Tree 完全二叉树的顺序存储,BST的性质 1066 Root of AVL Tree 构建AVL树,模板题,需理解记忆 1079 Total Sales of Supply Chain…

PAT甲级题目:点这里 pat解题列表 题号 标题 题目类型  10001 1001 A+B Format (20 分)  字符串处理  1003 1003 Emergency (25 分) 最短路径(Dijkstra or spfa)  1013 1013 Battle Over Cities (25 分) 图的遍历or并查集  1018 1018 Public Bike Management (30 分) 最短经 and 图的遍历       1030 1030 Travel Plan (3…

浏览全部代码:请戳 本文谨代表个人思路,欢迎讨论;) 1051. Pop Sequence (25) 题意 给定 stack 的容量,给定数据的入栈顺序:从 1 开始的正整数序列,在允许随机的出栈操作的情况下,要求判断某出栈序列是否可能. 比如,告知 stack 容量为 5,入栈序列的最大值为 7.有两个序列需要判断合理性: {1 2 3 4 5 6 7}: 这个序列是可能的,只需每次入栈时都做出栈操作. {3 2 1 7 5 6 4}: 这个序列是不可能的,其中前半部分 3 2 1 是合法的,…

最短路径 Emergency (25)-PAT甲级真题(Dijkstra算法) Public Bike Management (30)-PAT甲级真题(Dijkstra + DFS) Travel Plan (30)-PAT甲级真题(Dijkstra + DFS,输出路径,边权) All Roads Lead to Rome (30)-PAT甲级真题-Dijkstra + DFS Online Map (30)-PAT甲级真题(Dijkstra + DFS) 最短路径扩展问题 要求数最短路径有多…

本文为PAT甲级分类汇编系列文章. AVL树好难!(其实还好啦~) 我本来想着今天应该做不完树了,没想到电脑里有一份讲义,PPT和源代码都有,就一遍复习一遍抄码了一遍,更没想到的是编译一遍通过,再没想到的是运行也正常,最没想到的是一遍AC. 其实很多题都有数,std::set 之类用的是红黑树,据说很复杂,比AVL树还要复杂的那种.但是,用到这些设施的题,都不在这一分类下,这一分类下的题,因为题目要求自己建树,也就不用标准库设施了. 大多数题中,树在内存中都是连续存放的.不是像完全二叉树那样的连…