1053. Path of Equal Weight (30)

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

2.使用DFS进行遍历搜索（节点最多100个），能够满足要求

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;
}