PAT甲级——A1053 Path of Equal Weight

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, the number of nodes in a tree, M (<), the number of non-leaf nodes, and 0, the given weight number. The next line contains N positive numbers where W​i​​ (<) 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 { is said to be greater than sequence { if there exists 1 such that A​i​​=B​i​​ for ,, 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

深度遍历

 #include <iostream>
 #include <vector>
 #include <algorithm>
 using namespace std;
 struct Node
 {
     int val;
     vector<int>child;
 }node[];
 int N, M, S;
 int path[];
 void DFS(int head, int numNode, int sum)
 {
     if (sum > S)
         return;
     if (sum == S)
     {
         if (node[head].child.size() != )//不是叶子节点
             return;
         for (int i = ; i < numNode; ++i)
             cout << node[path[i]].val << (i < numNode -  ? " " : "");
         cout << endl;
         return;
     }
     for (int i = ; i < node[head].child.size(); ++i)
     {
         path[numNode] = node[head].child[i];
         DFS(node[head].child[i], numNode + , sum + node[node[head].child[i]].val);
     }
 }
 int main()
 {
     cin >> N >> M >> S;
     for (int i = ; i < N; ++i)
         cin >> node[i].val;
     int a, b, k;
     for (int i = ; i < M; ++i)
     {
         cin >> a >> k;
         for (int j = ; j < k; ++j)
         {
             cin >> b;
             node[a].child.push_back(b);
         }
         sort(node[a].child.begin(), node[a].child.end(),
             [](int a, int b) {return node[a].val > node[b].val; });
     }
     path[] = ;
     DFS(, , node[].val);
     return ;
 }