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

  1. 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 ...

  2. 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 W​i​​ assigne ...

  3. 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 ...

  4. 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 ...

  5. 1053 Path of Equal Weight (30分)(并查集)

    Given a non-empty tree with root R, and with weight W​i​​ assigned to each tree node T​i​​. The weig ...

  6. 【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 ...

  7. PAT (Advanced Level) 1053. Path of Equal Weight (30)

    简单DFS #include<cstdio> #include<cstring> #include<cmath> #include<vector> #i ...

  8. 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 ...

  9. 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

  1. Django实现Rbac权限管理

    权限管理 权限管理是根据不同的用户有相应的权限功能,通常用到的权限管理理念Rbac. Rbac 基于角色的权限访问控制(Role-Based Access Control)作为传统访问控制(自主访问, ...

  2. centos 7创建ss服务(方式二)

    一:安装pip yum install python-pip 如果没有python包则执行命令:yum -y install epel-release: 二:安装SS pip install shad ...

  3. 洛谷P1038神经网络题解

    题目 这个题不得不说是一道大坑题,为什么这么说呢,这题目不仅难懂,还非常适合那种被生物奥赛刷下来而来到信息奥赛的学生. 因此我们先分析一下题目的坑点. 1: 题目的图分为输入层,输出层,以及中间层. ...

  4. atcoder NIKKEI Programming Contest 2019 E - Weights on Vertices and Edges

    题目链接:Weights on Vertices and Edges 题目大意:有一个\(n\)个点\(m\)条边的无向图,点有点权,边有边权,问至少删去多少条边使得对于剩下的每一条边,它所在的联通块 ...

  5. nginx+php使用open_basedir限制站点目录防止跨站

    以下三种设置方法均需要PHP版本为5.3或者以上.方法1)在Nginx配置文件中加入 fastcgi_param PHP_VALUE "open_basedir=$document_root ...

  6. python学习日记(装饰器的补充)

    如何返回被装饰函数的函数名及注释? 问题及实现 先看典型的装饰器: def wrapper(f):#装饰器函数,f是被装饰函数 def inner(*args,**kwargs): '''执行函数之前 ...

  7. Configure new Nagios clients

    安装rpm -Uvh http://dl.fedoraproject.org/pub/epel/6/x86_64/epel-release-6-8.noarch.rpmrpm -Uvh http:// ...

  8. 计算几何细节梳理&模板

    点击%XZY巨佬 向量的板子 #include<bits/stdc++.h> #define I inline using namespace std; typedef double DB ...

  9. [luogu1110][ZJOI2007]报表统计【平衡树】

    传送门 [洛谷传送门] [bzoj传送门] 前言 洛谷和网上的题解都好复杂哦,或者是stl水过. 窝的语文不怎么好,所以会有一些表达上的累赘或者是含糊不清,望各大佬海涵. 前置芝士 首先你一定要会平衡 ...

  10. X-PACK详解

    启用和禁用启用和禁用X-Pack功能默认情况下,所有X-Pack功能都被启用.您可以启用或禁用特定的X-Pack功能elasticsearch.yml,kibana.yml以及logstash.yml ...