PAT 1053 Path of Equal Weight[比较]】的更多相关文章

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…
#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…
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…
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,…
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<…