1053. Path of Equal Weight (30)
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 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 = Bifor 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
#include<iostream>
#include<vector>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
#define max 102
vector<int>vt[max];
vector<int>result[max];
vector<int>weight;
int sweight;
int t=0;
vector<int>curpath;
void dfs(int s,int len){
if(vt[s].empty())return;
int tmplen = len;
int size=vt[s].size();
for(int i=0;i<size;i++){
tmplen=len + weight[vt[s][i]];
if(tmplen<sweight){
vector<int>tmp;
tmp=curpath;
curpath.push_back(weight[vt[s][i]]);
dfs(vt[s][i],tmplen);
curpath=tmp;
}else if(tmplen==sweight && vt[vt[s][i]].empty()){
result[t]=curpath;
result[t].push_back(weight[vt[s][i]]);
t++;
}
}
}
bool cmp(vector<int>a,vector<int>b){
int sizea=a.size();
int sizeb=b.size();
int minSize=sizea<sizeb?sizea:sizeb;
for(int i=0;i<minSize;i++){
if(a[i]>b[i])return true;
else if(a[i]<b[i]) return false;
}
if(sizea==minSize){
return false;
}else {
return true;
}
}
int main(){
int n,m;
scanf("%d%d%d",&n,&m,&sweight);
int i,j;
int val,id,k;
weight.resize(n);
for(i=0;i<n;i++){
scanf("%d",&weight[i]);
}
for(i=0;i<m;i++){
scanf("%d%d",&id,&k);
for(j=0;j<k;j++){
scanf("%d",&val);
vt[id].push_back(val);
}
}
if(m==0){
if(weight[0]==sweight)printf("%d\n",weight[0]);
return 0;
}
curpath.push_back(weight[0]);
dfs(0,weight[0]);
sort(result,result+t,cmp);
for(i=0;i<t;i++){
int size=result[i].size();
printf("%d",result[i][0]);
for(j=1;j<size;j++){
printf(" %d",result[i][j]);
}
printf("\n");
}
return 0;
}
1053. Path of Equal Weight (30)的更多相关文章
- 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 ...
- 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 ...
- 1053 Path of Equal Weight (30)(30 分)
Given a non-empty tree with root R, and with weight W~i~ assigned to each tree node T~i~. The weight ...
- PAT Advanced 1053 Path of Equal Weight (30) [树的遍历]
题目 Given a non-empty tree with root R, and with weight Wi assigned to each tree node Ti. The weight ...
- 1053 Path of Equal Weight (30分)(并查集)
Given a non-empty tree with root R, and with weight Wi assigned to each tree node Ti. The weig ...
- 【PAT甲级】1053 Path of Equal Weight (30 分)(DFS)
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 ...
- PAT (Advanced Level) 1053. Path of Equal Weight (30)
简单DFS #include<cstdio> #include<cstring> #include<cmath> #include<vector> #i ...
- 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 ...
- 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 ...
随机推荐
- [离散时间信号处理学习笔记] 3. 一些基本的LTI系统
首先我们需要先对离散时间系统进行概念上的回顾: $y[n] = T\{ x[n] \}$ 上面的式子表征了离散时间系统,也就是把输入序列$x[n]$,映射称为$y[n]$的输出序列. 不过上述式子也可 ...
- 搭建Hexo博客(三)—换电脑继续写Hexo博客
Hexo和GitHub搭建博客的原理是:Hexo将source下的md文件生成静态的html页面,存放到public目录中,这一步是由命令:hexo -g完成.接下来执行hexo -d命令,就将pub ...
- Jarvis OJ A Piece Of Cake
看图片的隐写术自闭,本来想看一看jarvisoj 的basic放松一下心情,结果一道题就做了一晚上qwq 首先看到这道题的时候想到的是凯撒密码(这其实是Google之后才知道这个名字的)枚举了26种位 ...
- NAND闪存供过于求的情况今年会有所好转吗?
2018年,NAND闪存全年供过于求,价格一直下跌,导致西数.东芝等厂商毛利率大幅下滑.如今到了2019年,情况会有所好转吗? 近日,集邦科技旗下半导体研究中心DRAMeXchange发布调查报告指出 ...
- Ubuntu18.04安装RabbitMQ
Ubuntu18.04安装RabbitMQ 2018年06月10日 19:32:38 dmfrm 阅读数:2492 版权声明:本文为博主原创文章,未经博主允许不得转载. https://blog ...
- C#常忘语法笔记(C#程序设计基础1-4章)
1.1 const:声明一个常量 1.2强转: double->int eg1: int i=(int)3.0; eg2: double d=3.0; int i=(int)d+1; strin ...
- Spring MVC 使用介绍(二)—— DispatcherServlet
一.Hello World示例 1.引入依赖 <dependency> <groupId>javax.servlet</groupId> <artifactI ...
- Github提交本地代码
第一步:建立git仓库 cd到你的本地项目根目录下,执行git命令 git init 第二步:将项目的所有文件添加到仓库中 git add . 如果想添加某个特定的文件,只需把.换成特定的文件名即可 ...
- linux系统版本大全
Linux系统下载地址:http://www.jb51.net/LINUXjishu/239493.html linux系统教学视频:http://www.uplinux.com/shipin/lin ...
- win10安装MySql 5.7.23
下载安装 因为Django2.1不再支持MySQL5.5,这里需要重新安装一下MySQL 首先去官网下载 这里使用的是msi版本 https://dev.mysql.com/downloads/win ...