# 1053. Path of Equal Weight (30)

dfs函数携带vector形参记录搜索路径

10 ms

65536 kB

16000 B

Standard

CHEN, Yue

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 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 1

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 ID ... 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 = Bifor 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```
```
`#include <iostream>`
`#include <map>`
`#include <vector>`
`#include <algorithm>`

`using namespace std;`

`map<int, vector<int>> mp;`

`vector<int> weight;`
`int n, m, aimWeight;`
`int pathWeight=0;`

`bool cmp(int a, int b) {`
`	return weight[a] > weight[b];`
`}`

`void dfs(int i,int w, vector<int> p) {`
`	p.push_back(i);`

`	if (mp[i].size() == 0) {`
`		if (w == aimWeight) {`
`			for (int j = 0; j < p.size(); j++) {`
`				cout << weight[p[j]];`
`				if (j != p.size() - 1)`
`					cout << " ";`
`				else`
`					cout << endl;`
`			}`
`		}`
`		p.clear();`
`		return;`
`	}`

`	else {`
`		for (int j = 0; j < mp[i].size(); j++)`
`			dfs(mp[i][j],weight[mp[i][j]]+w,p);`
`	}`
`}`

`int main(void) {	`
`	cin >> n >> m >> aimWeight;`
`	for (int i = 0; i < n; i++) {`
`		int wtemp;`
`		cin >> wtemp;`
`		weight.push_back(wtemp);`
`	}`
`	for (int i = 0; i < m; i++) {`
`		int id, k;`
`		cin >> id >> k;`
`		for (int j = 0; j < k; j++) {`
`			int idtemp;`
`			cin >> idtemp;`
`			mp[id].push_back(idtemp);`
`		}`
`		sort(mp[id].begin(), mp[id].end(), cmp);`
`	}`
`	vector<int> path;`
`	dfs(0,weight,path);`

`	return 0;`
`}`

```

## PAT：1053. Path of Equal Weight (30) AC

#include<stdio.h> #include<vector> #include<queue> #include<algorithm> using namespace std; const int MAX=1010; int n,m; //n个节点,m个非叶子节点 long long int S; //待测权值 long long int weight[MAX]; //每个节点的权值 vector<int> child[MAX]; //存储

## PAT甲题题解-1053. Path of Equal Weight (30)-dfs ## 1053 Path of Equal Weight (30分) 1. 题目 2. 思路 定义结构体, 并且使用下标作为序号 struct node{ string weight; vector<int> children; }nodes; 读取数据,并且排序children,方便输出 使用先序遍历,处理数据 3. 注意点 权重的值很大,用字符串处理,要自己写加法和比较函数 4. 代码 #include<cstdio> #include<algorithm> #include<vector> #include&l

## PAT Advanced Level 1053 Path of Equal Weight

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 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