1053. Path of Equal Weight (30)

1.题目给出一棵和一个权重值W，要求求出从根到叶子节点权重累加为W的路径。

2.该题使用树节点数据结构，其中包含vector<int> child列表和权重值。

3.使用DFS进行遍历搜索（节点最多100个），能够满足要求。

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 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 1Input 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 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, A₂, …, A_n} is said to be greater than sequence {B₁, B₂, …, B_m} if there exists 1 <= k < min{n, m} such that A_i = B_ifor 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

 
AC代码：

//#include<string>
//#include<stack>
//#include<unordered_set>
//#include <sstream>
//#include "func.h"
//#include <list>
#include <iomanip>
#include<unordered_map>
#include<set>
#include<queue>
#include<map>
#include<vector>
#include <algorithm>
#include<stdio.h>
#include<iostream>
#include<string>
#include<memory.h>
#include<limits.h>
#include<stack>
using namespace std;
struct TreeNode{
	vector<int> child;
	int w;
	TreeNode() :child(0), w(0){};
};
void dfs(int now, int target, vector<TreeNode>&tree, vector<int>&path, vector<vector<int>>&ans)
{
	if (tree[now].child.size() == 0 && target == 0)
		ans.push_back(path);//叶子节点，并且满足要求
	if (tree[now].child.size() != 0 && target <= 0)
		return;//
	else if (target < 0)
		return;
	for (int i = 0; i < tree[now].child.size(); i++)
	{
		path.push_back(tree[tree[now].child[i]].w);
		dfs(tree[now].child[i], target - tree[tree[now].child[i]].w, tree, path, ans);
		path.pop_back();
	}
}
bool cmp(const vector<int>&a, const vector<int>&b)
{
	for (int i = 0; i < min(a.size(), b.size()); i++)
	{
		if (a[i] >b[i])
			return true;
		else if (a[i] < b[i])
			return false;
	}
	return false;
}
int main(void)
{
	int nodeSum, nonLeafSum, target;
	cin >> nodeSum >> nonLeafSum >> target;
	vector<TreeNode> tree(nodeSum);
	
	for (int i = 0; i < nodeSum; i++)
	{
		scanf("%d", &tree[i].w);
	}

	for (int i = 0; i < nonLeafSum; i++)
	{
		int nodeNum=0;
		scanf("%d", &nodeNum);
		int childSum = 0;
		scanf("%d", &childSum);
		tree[nodeNum].child = vector<int>(childSum);
		for (int j = 0; j < childSum; j++)
		{
			scanf("%d", &tree[nodeNum].child[j]);
		}
	}
	if (nodeSum > 0)
	{
		vector<int> path;
		vector<vector<int>> ans;
		path.push_back(tree[0].w);//注意判断根节点是否存在
		target -= tree[0].w;
		dfs(0, target, tree, path, ans);
		sort(ans.begin(), ans.end(), cmp);
		for (int i = 0; i < ans.size(); i++)
		{
			for (int j = 0; j < ans[i].size(); j++)
			{
				printf("%d", ans[i][j]);
				if (j != ans[i].size() - 1)
					printf(" ");
			}
			printf("\n");
		}

	}
	return 0;
}