PAT A 甲級 1053 Path of Equal Weight (30分)

1053 Path of Equal Weight (30分)

吐槽

這什麼莫名其妙的輸出排序要求啊喂！我實作這個排序的用時比完成題目主幹内容還長……

題意

有一棵樹，每個節點有重量（或者其實是“權值”？），從根到一個葉稱作一條路徑，路徑的重量為路徑上各節點重量之和，求所有重量為要求值的路徑并将他們按同層節點重量的從大到小排列

思路

其實感覺題目沒什麼難度啊。因為每條路徑是從根到葉，而每個節點的父節點有且隻有一個，是以記錄父節點比記錄子節點更高效——最後把每個葉節點路徑向上追溯，算出重量，即可知道哪些路徑合法。

*如果是從根到葉遞歸計算的操作，似乎會有靠動态規劃進而快于本算法的餘地，但是這道題的資料量不需要動态規劃。（說實話我覺得需要動态規劃的話才更符合它30分的分值）

至于排序……反正就是sort函數+cmp函數解決。（我用時長是因為本來的stack+及時輸出的結構被迫改成了vector+儲存、排序後輸出）

新學習

在sort使用的cmp函數中，true和false代表的是“需要變化位置”和“不需要變化位置”。是以要比較的兩個東西相等時應該return false。

代碼

#include<iostream>
#include<stack>
#include<vector>
#include<algorithm>

using namespace std;

bool cmp(vector<int> v1, vector<int> v2)
{
	int i = 0;
	while (i < v1.size() && i < v2.size())
	{
		if (v1[i] < v2[i])
			return false;
		else if (v1[i] > v2[i])
			return true;
		i++;
	}
	return false;
}

typedef struct
{
	int weight;
	int father;
	bool leaf = true;
}Node;

int main()
{
	int n, m, s, i, ii, temp, temp2, temp3;
	cin >> n >> m >> s;
	Node* node;
	node = new Node[n];
	vector<vector<int>> vv;
	vector<int> v;
	for (i = 0; i < n; i++)
	{
		cin >> node[i].weight;
	}
	for (i = 0; i < m; i++)
	{
		cin >> temp >> temp2;
		node[temp].leaf = false;
		for (ii = 0; ii < temp2; ii++)
		{
			cin >> temp3;
			node[temp3].father = temp;
		}
	}
	for (i = 0; i < n; i++)
	{
		if (!node[i].leaf)
			continue;
		temp = i;
		temp2 = 0;
		while (temp != 0 && temp2 < s)
		{
			temp2 += node[temp].weight;
			temp = node[temp].father;
		}
		temp2 += node[0].weight;
		if (temp2 == s)
		{
			v.clear();
			temp = i;
			while (temp != 0)
			{
				v.push_back(node[temp].weight);
				temp = node[temp].father;
			}
			reverse(v.begin(), v.end());
			vv.push_back(v);
		}
	}

	sort(vv.begin(), vv.end(), cmp);
	for (i = 0; i < vv.size(); i++)
	{
		cout << node[0].weight;
		for (ii = 0; ii < vv[i].size(); ii++)
		{
			cout << " " << vv[i][ii];
		}
		cout << endl;
	}
	return 0;
}
           

題目

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 {A1,A​2,⋯,An} is said to be greater than sequence {B​1,B​2,⋯,Bm} if there exists 1≤k<min{n,m} such that Ai=B​i for i=1,⋯,k, and A​k+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