【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 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 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<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
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
- 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 ...
- 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)
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 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 ...
- 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 ...
- 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 ...
- 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
- windows10中远程登录ubuntu16.04的桌面
1. 安装xrdp sudo apt-get install xrdp 2. 安装vnc4server sudo apt-get install vnc4server tightvncserver ...
- Could not load file or assembly 'Microsoft.ReportViewer.WebForms, Version=11.0.0.0, Culture=neutral, PublicKeyToken=89845dcd8080cc91' or one of its dependencies
my shurufa huai diao le 1\ first you need install " SQLSysClrTypes_x86.msi " 2\ ...
- 洛谷P3928 Sequence2(dp,线段树)
题目链接: 洛谷 题目大意在描述底下有.此处不赘述. 明显是个类似于LIS的dp. 令 $dp[i][j]$ 表示: $j=1$ 时表示已经处理了 $i$ 个数,上一个选的数来自序列 $A[0]$ 的 ...
- HDU 4280 Island Transport(网络流,最大流)
HDU 4280 Island Transport(网络流,最大流) Description In the vast waters far far away, there are many islan ...
- sqlite3数据库的简要应用
Sqlite3数据库升级方案的变化. 1, 若是讲要升级的数据库版本更高,则从低版本数据库中拷贝与新数据库相同字段的内容,其他字段按照默认值创建.A->B->C这样逐个版本升级,每个版本 ...
- gdb调试2—单步执行和跟踪函数
int add_range(int low, int high); int main(int argc, char *argv[]) { int result[100]; result[0] = ad ...
- LVS NAT/DR
LVS介绍:http://www.linuxvirtualserver.org/zh/lvs3.html DR 工作流: HOST发送服务请求报文(源IP为HOST IP,目的IP为VSIP) Gen ...
- Tomcat权威指南-读书摘要系列10
Tomcat集群 一些集群技术 DNS请求分配 TCP网络地址转换请求分配 Mod_proxy_balance负载均衡与故障复原 JDBC请求分布与故障复原
- Java基础-通过POI接口处理xls
Java基础-通过POI接口处理xls 作者:尹正杰 版权声明:原创作品,谢绝转载!否则将追究法律责任.
- [转]Android ANR 分析解决方法
一:什么是ANR ANR:Application Not Responding,即应用无响应 二:ANR的类型 ANR一般有三种类型: 1. KeyDispatchTimeout(5 seconds) ...