PAT甲级真题 1053 Path of Equal Weight (30分) C++实现（DFS遍历树）

该博客介绍了如何解决PAT甲级竞赛中的一道问题，即寻找一棵给定权重的树中所有权重等于特定值的路径。作者提供了一种使用DFS遍历树的方法，并详细解释了思路，包括利用二维数组存储树结构，按权重排序孩子节点，以及从根节点开始的深度优先搜索策略。文章包含输入输出规格，样例输入和输出，以及具体实现代码。

题目

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.
在这里插入图片描述
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[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, …, An} is said to be greater than sequence {B1, B2, …, Bm} if there exists 1 <= k < min{n, m} such that Ai = Bi for 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:

思路

用二维数组vector<vector<int> > child记录树，child[i]表示编号为i的节点的孩子数组。将孩子按权重从大到小排序，这样顺序dfs遍历孩子数组即可得到降序路径。

从根节点执行dfs，记录当前的层数layer和权重和sum。以当前层数为索引记录遍历路径path，当找到一条路径时（到达叶子节点且sum==s），顺序输出当前path上节点对应的权重即可。

代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int n, m, s;
vector<int> w;
vector<int> path(100);
vector<vector<int> > child(100);

bool cmp(int a, int b){
    return w[a] > w[b];
}
void dfs(int root, int layer, int sum){
    sum += w[root];
    path[layer] = root;
    int k = child[root].size();
    //到达叶子节点
    if (k==0){
        if (sum==s){
            cout << w[0];
            for (int i=1; i<=layer; i++){
                cout << " " << w[path[i]];
            }
            cout << endl;
        }
    }
    else{
        for (int i=0; i<k; i++){
            dfs(child[root][i], layer+1, sum);
        }
    }
}

int main(){
    cin >> n >> m >> s;
    w.resize(n);
    for (int i=0; i<n; i++){
        cin >> w[i];
    }
    for (int i=0; i<m; i++){
        int root, k;
        cin >> root >> k;
        child[root].resize(k);
        for (int j=0; j<k; j++){
            cin >> child[root][j];
        }
        sort(child[root].begin(), child[root].end(), cmp);
    }
    dfs(0, 0, 0);
    return 0;
}