PAT甲级——A1053 Path of Equal Weight
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 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 Wi (<) 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 { is said to be greater than sequence { if there exists 1 such that Ai=Bi for ,, 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
深度遍历
1 #include <iostream> 2 #include <vector> 3 #include <algorithm> 4 using namespace std; 5 struct Node 6 { 7 int val; 8 vector<int>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 }