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 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 the following figure: 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 the figure.

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<2​30​​, the given weight number. The next line contains N positive numbers where W​i​​ (<1000) corresponds to the tree node T​i​​. 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 {A​1​​,A​2​​,⋯,A​n​​} is said to be greater than sequence {B​1​​,B​2​​,⋯,B​m​​} if there exists 1≤k<min{n,m} such that A​i​​=B​i​​ for i=1,⋯,k, and A​k+1​​>B​k+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

C++代码如下:

 #include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
#define maxn 105 struct Node {
int weight;
vector<int>child;
}; int n, m, s;
Node num[maxn]; bool cmp(int a, int b) {
return num[a].weight > num[b].weight;
} vector<int>v; //存放路径对应的权值
void path(int r,int sum) {
if (sum > s) {
v.pop_back(); return;
}
if (sum == s) {
if ( num[r].child.size() == ) {
cout << v[];
for (vector<int>::iterator it = v.begin() + ; it != v.end(); it++)
cout << ' ' << *it;
cout << endl;
v.pop_back();
return;
}
else {
v.pop_back(); return;
}
}
for (int i = ; i < num[r].child.size(); i++) {
int t = num[r].child[i];
v.push_back(num[t].weight);
path(t, sum + num[t].weight);
}
if (!v.empty()) v.pop_back();
}
int main() {
cin >> n >> m >> s;
int w; for (int i = ; i < n; i++) {
cin >> w;
num[i].weight = w;
}
int id,k,t;
for (int i = ; i < m; i++) {
cin >> id>>k;
for (int j = ; j < k; j++) {
cin >> t;
num[id].child.push_back(t);
}
sort(num[id].child.begin(), num[id].child.end(), cmp);
}
v.push_back(num[].weight);
path(,num[].weight);
return ;
}

【PAT】1053 Path of Equal Weight(30 分)的更多相关文章

  1. 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 ...

  2. 【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 ...

  3. 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 ...

  4. 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 ...

  5. 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 W​i​​ assigned t ...

  6. 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 ...

  7. 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 ...

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

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

  9. 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 ...

  10. 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 ...

随机推荐

  1. BZOJ2303 APIO2011方格染色(并查集)

    比较难想到的是将题目中的要求看做异或.那么有ai,j^ai+1,j^ai,j+1^ai+1,j+1=1.瞎化一化可以大胆猜想得到a1,1^a1,j^ai,1^ai,j=(i-1)*(j-1)& ...

  2. Vasya and Multisets CodeForces - 1051C(英语限制了我的想象力)

    题意: 对输入的每个数都进行分配到两个多重集中,问分配完以后 每个多重集中的个数为1的元素是否相等  相等则输出一个分配方式, 不想等为NO 解析: 三种情况 1.原数列中个数为1 的数的个数为偶数 ...

  3. 骑士 HYSBZ - 1040(基环树+树形dp)

    Z国的骑士团是一个很有势力的组织,帮会中汇聚了来自各地的精英.他们劫富济贫,惩恶扬善,受到社会各界的赞扬.最近发生了一件可怕的事情,邪恶的Y国发动了一场针对Z国的侵略战争.战火绵延五百里,在和平环境中 ...

  4. 为harbor部署swagger-ui

    有了swagger,所有REST API就一览无余了. Harbor官方已经准备好了部署脚本,部署起来其实特别简单,这里仅作记录. 1.下载部署脚本和swagger文件到本地harbor目录. # c ...

  5. activiti复盘重推的一种简单实现方式:

    activiti复盘重推的一种简单实现方式: 设置流程的每一步让用户选择,比如一共有6步完成,用户选择从第4步开始复盘重推,那么把原来的推演oldId和4传到后台, 首先,后台生成一个新的推演id n ...

  6. 基于CMS的组件复用实践

    目前前端项目大多基于Vue.React.Angular等框架来实现,这一类框架都有一个明显的特点:基于模块化以及组件化思维.所以,开发者在使用上述框架时,实际上是在写一个一个的组件,并且组件与组件之间 ...

  7. k8s pod的4种网络模式最佳实战(externalIPs )

    [k8s]k8s pod的4种网络模式最佳实战(externalIPs )       hostPort相当于docker run -p 8081:8080,不用创建svc,因此端口只在容器运行的vm ...

  8. SqlServer在视图上创建索引

    在视图上创建索引需要三个条件: 一.视图必须绑定到架构. 要做到这点,在 CREATE VIEW 语句中,必须加上 WITH SCHEMABINDING,如果是使用企业管理器,则在设计界面的空白处点击 ...

  9. shell jq

    Mark 下,周末来补充 参考资料: https://stedolan.github.io/jq/tutorial/

  10. 设计模式之Mixin模式

    介绍 mixin模式就是一些提供能够被一个或者一组子类简单继承功能的类,意在重用其功能.在面向对象的语言中,我们会通过接口继承的方式来实现功能的复用.但是在javascript中,我们没办法通过接口继 ...