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…

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(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 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…

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 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…

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…

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…

#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…

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…