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. 51Nod 1199 Money out of Thin Air (树链剖分+线段树)

    1199 Money out of Thin Air  题目来源: Ural 基准时间限制:1 秒 空间限制:131072 KB 分值: 80 难度:5级算法题  收藏  关注 一棵有N个节点的树,每 ...

  2. 【转】如何向XML内插入一个字符串片段

    转自:http://bbs.csdn.net/topics/190051229   5楼 string filepath = Server.MapPath("你的xml文件"); ...

  3. 【BZOJ1294】[SCOI2009]围豆豆(动态规划,状压)

    [BZOJ1294][SCOI2009]围豆豆(动态规划,状压) 题面 BZOJ 洛谷 题解 首先考虑如何判断一个点是否在一个多边形内(不一定是凸的),我们从这个点开始,朝着一个方向画一条射线,看看它 ...

  4. POJ 1502 MPI Maelstrom / UVA 432 MPI Maelstrom / SCU 1068 MPI Maelstrom / UVALive 5398 MPI Maelstrom /ZOJ 1291 MPI Maelstrom (最短路径)

    POJ 1502 MPI Maelstrom / UVA 432 MPI Maelstrom / SCU 1068 MPI Maelstrom / UVALive 5398 MPI Maelstrom ...

  5. 《剑指offer》— JavaScript(33)丑数

    丑数 题目描述 把只包含质因子2.3和5的数称作丑数(Ugly Number).例如6.8都是丑数,但14不是,因为它包含因子7. 习惯上我们把1当做是第一个丑数.求按从小到大的顺序的第N个丑数. 思 ...

  6. Spring中Model,ModelMap以及ModelAndView之间的区别

    原文链接:http://blog.csdn.net/zhangxing52077/article/details/75193948 Spring中Model,ModelMap以及ModelAndVie ...

  7. JVM调优命令-jstat

    JVM Statistics Monitoring Tool,是用于监视虚拟机运行时状态信息的命令,它可以显示出虚拟机进程中的类装载.内存.垃圾收集.JIT编译等运行数据.[性能分析] 命令格式 1 ...

  8. keepalived 配置文件参数详解

    global_defs 全局配置vrrpd 1. vrrp_script添加一个周期性执行的脚本.脚本的退出状态码会被调用它的所有的VRRP Instance记录. 2. vrrp_sync_grou ...

  9. bzoj千题计划207:bzoj1879: [Sdoi2009]Bill的挑战

    http://www.lydsy.com/JudgeOnline/problem.php?id=1879 f[i][j] 表示匹配了i个字符,匹配字符串的状态为j的方案数 枚举下一个字符是什么 计算加 ...

  10. CF&&CC百套计划4 Codeforces Round #276 (Div. 1) A. Bits

    http://codeforces.com/contest/484/problem/A 题意: 询问[a,b]中二进制位1最多且最小的数 贪心,假设开始每一位都是1 从高位i开始枚举, 如果当前数&g ...