Path of Equal Weight (DFS)
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 given any weighted tree, you are supposed to find all the paths with their weights equal to a given number. For example, let's consider the tree showed
in Figure 1: for each node, the upper number is the node ID which is a two-digit number, and the lower number is the weight of that node. Suppose that the given number is 24, then there exists 4 different paths which have the same given weight: {10 5 2 7},
{10 4 10}, {10 3 3 6 2} and {10 3 3 6 2}, which correspond to the red edges in Figure 1.
Figure 1
Input Specification:
Each input file contains one test case. Each case starts with a line containing 0 < N <= 100, the number of nodes in a tree, M (< N), the number of non-leaf
nodes, and 0 < S < 230, the given weight number. The next line contains N positive numbers where Wi (<1000) corresponds to the tree node Ti. Then M lines follow, each in the format:
ID K ID[1] ID[2] ... ID[K]
where ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID's of its children.
For the sake of simplicity, let us fix the root ID to be 00.
Output Specification:
For each test case, print all the paths with weight S in non-increasing order. Each path occupies a line with printed weights from the root to the leaf in
order. All the numbers must be separated by a space with no extra space at the end of the line.
Note: sequence {A1, A2, ..., An} is said to be greater than sequence {B1, B2, ..., Bm} if there exists 1 <= k < min{n, m} such that Ai = Bi for i=1, ... k,
and Ak+1 > Bk+1.
Sample Input:
20 9 24
10 2 4 3 5 10 2 18 9 7 2 2 1 3 12 1 8 6 2 2
00 4 01 02 03 04
02 1 05
04 2 06 07
03 3 11 12 13
06 1 09
07 2 08 10
16 1 15
13 3 14 16 17
17 2 18 19
Sample Output:
10 5 2 7
10 4 10
10 3 3 6 2
10 3 3 6 2
这道30分的题目,提交一次就意外的AC了。
就是 建立连接表 DFS+记录路径+权值累加 搜到叶子节点,如果权值之和与要求的的相等时保存路径。
最后的排序要点混,但进行三层的判断排序,也就能过了,
#include <iostream> #include <string> #include <vector> #include <algorithm> using namespace std; int WW[100]; int visit[100]; vector<int> vv[100]; vector<int> road; vector<int> RR[100]; int sum,wi; bool cmp(vector<int> a,vector<int> b) { if(a[0]==b[0]&&a[1]==b[1]) return a[2]>b[2]; if(a[0]==b[0]) return a[1]>b[1]; return a[0]>b[0]; } void DFS(int root,int &count) { if(visit[root]==0) { visit[root]=1; road.push_back(WW[root]); sum+=WW[root]; for(int i=0;i<vv[root].size();i++) { if(visit[vv[root][i]]==0) DFS(vv[root][i],count); } if(sum==wi&&vv[root].size()==0) { RR[count++]=road; } road.pop_back(); sum-=WW[root]; } } int main() { int i,j,num,fnum; while(cin>>num) { road.clear(); cin>>fnum>>wi; for(i=0;i<num;i++) { cin>>WW[i]; vv[i].clear(); visit[i]=0; RR[i].clear(); } for(i=0;i<fnum;i++) { int n1,n2; cin>>n1>>n2; for(j=0;j<n2;j++) { int tem; cin>>tem; vv[n1].push_back(tem); } } int count=0; sum=0; DFS(0,count); sort(RR,RR+count,cmp); for(i=0;i<count;i++) { cout<<RR[i][0]; for(j=1;j<RR[i].size();j++) cout<<" "<<RR[i][j]; cout<<endl; } } return 0; }
Path of Equal Weight (DFS)的更多相关文章
- PAT 甲级 1053 Path of Equal Weight (30 分)(dfs,vector内元素排序,有一小坑点)
1053 Path of Equal Weight (30 分) Given a non-empty tree with root R, and with weight Wi assigne ...
- PAT 1053 Path of Equal Weight[比较]
1053 Path of Equal Weight(30 分) Given a non-empty tree with root R, and with weight Wi assigned t ...
- pat1053. Path of Equal Weight (30)
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 G ...
- pat 甲级 1053. Path of Equal Weight (30)
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 ...
- 1053 Path of Equal Weight——PAT甲级真题
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 L ...
- 【PAT】1053 Path of Equal Weight(30 分)
1053 Path of Equal Weight(30 分) Given a non-empty tree with root R, and with weight Wi assigned t ...
- PAT_A1053#Path of Equal Weight
Source: PAT A1053 Path of Equal Weight (30 分) Description: Given a non-empty tree with root R, and w ...
- PAT 1053 Path of Equal Weight
#include <cstdio> #include <cstdlib> #include <vector> #include <algorithm> ...
- PAT甲题题解-1053. Path of Equal Weight (30)-dfs
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 ...
随机推荐
- PHP读书笔记(5)-结构语句
PHP结构语句 顺序结构 顺序结构就像一条直线,按着顺序一直往下执行.我们编写的代码默认都是按照顺序结构执行的. 条件结构之if…else… 条件结构就像一个岔路口,可以向左走,也可以向右走.比如上洗 ...
- 关于JAVA那点事---i++和++i
对i++和++i 一直有点晕,今天专门抽空来研究相关的知识. 先从简单的说起. 有如下程序: int i=0; i=i++; system.out.print(i); 一眼望去这个结果显而易见是1,但 ...
- asp.net微信支付打通发货通知代码
上次遇到微信支付,发货接口的时候,官方的demo也没有提供相应的代码 ,因本人技术有限,百度 google 很久都没有asp.net 版本的,最后只好硬着头皮自己搞,没想到官方文档也是错的. 我这一步 ...
- chrome浏览器取消置顶的方法
这两天在使用google chrome浏览器的时候,发现chrome被默认置顶,取消chrome默认的方法为在浏览器上按 “ALT + Space + C”,然后再重开chorme就可以了.
- ActiveMQ(5.10.0) - Destination-level authorization
To build upon authentication, consider a use case requiring more fine-grained control over clients t ...
- 和阿文一起学H5--如何把H5压缩到最小
三种压缩图片的方法: 1.PS 但是PS每次只能压缩一张,下面介绍第二个神器 2.TinyPng压缩 https://tinypng.com/ 3.IloveIMG压缩 http://www.ilov ...
- Android之单选框
MainActivity继承Activity使用OnCheckedChangeListener接口: package com.cdp.checkbox; import android.app.Acti ...
- Oracle 一些操作
Achivelog ============================ alter system set db_recovery_file_dest='F:\ORACLE\recovery_ar ...
- 在swift中使用MJRefresh
cocoapod导入的,并且桥接已经完成,但是就是不提示方法,醉了,
- Android应用资源--之属性(Attribute)资源
原文链接: http://wujiandong.iteye.com/blog/1184921 属性(Attribute)资源:属于整个Android应用资源的一部分.其实就是网上一堆介绍怎么给自定义V ...