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 2410 2 4 3 5 10 2 18 9 7 2 2 1 3 12 1 8 6 2 200 4 01 02 03 0402 1 0504 2 06 0703 3 11 12 1306 1 0907 2 08 1016 1 1513 3 14 16 1717 2 18 19

Sample Output:

10 5 2 710 4 1010 3 3 6 210 3 3 6 2 深度遍历

1 #include 
     
       2 #include 
      
        3 #include 
       
         4 using namespace std; 5 struct Node 6 { 7     int val; 8     vector
        
         child; 9 }node[101];10 int N, M, S;11 int path[101];12 void DFS(int head, int numNode, int sum)13 {14     if (sum > S)15         return;16     if (sum == S)17     {18         if (node[head].child.size() != 0)//不是叶子节点19             return;20         for (int i = 0; i < numNode; ++i)21             cout << node[path[i]].val << (i < numNode - 1 ? " " : "");22         cout << endl;23         return;24     }25     for (int i = 0; i < node[head].child.size(); ++i)26     {27         path[numNode] = node[head].child[i];28         DFS(node[head].child[i], numNode + 1, sum + node[node[head].child[i]].val);29     }30 }31 int main()32 {33     cin >> N >> M >> S;34     for (int i = 0; i < N; ++i)35         cin >> node[i].val;36     int a, b, k;37     for (int i = 0; i < M; ++i)38     {39         cin >> a >> k;40         for (int j = 0; j < k; ++j)41         {42             cin >> b;43             node[a].child.push_back(b);44         }45         sort(node[a].child.begin(), node[a].child.end(),46             [](int a, int b) {
   return node[a].val > node[b].val; });47     }48     path[0] = 0;49     DFS(0, 1, node[0].val);50     return 0;51 }