PAT_1053: 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 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 W_i (<1000) 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 {A₁, A₂, ..., A_n} is said to be greater than sequence {B₁, B₂, ..., B_m} if there exists 1 <= k < min{n, m} such that A_i = B_ifor i=1, ... k, 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

备注： 从Root开始进行DFS遍历，遍历时记录path，遍历到leaf时进行和的check。最后对所有找到的path进行一个题目要求的排序即可。PS：sort函数可真好用

#include<iostream>#include<vector>#include<algorithm>using namespace std;typedef struct node{int weight;vector<int> child;}NODE;NODE tree[100];vector<vector<int>> path; //store all the resulting pathvector<int> current_path;void FindAllPath(int given_s, int node_id){NODE current_node=tree[node_id];if(current_node.child.size()==0) //it is a leaf-node, check the sum{int current_sum = 0;for(int i=0;i<(int)current_path.size();i++)current_sum+=tree[current_path[i]].weight;if(current_sum==given_s) //found a path!path.push_back(current_path);}else //non-leaf,doing a recursive call{for(int i=0;i<(int)current_node.child.size();i++){int child = current_node.child[i];current_path.push_back(child);FindAllPath(given_s,child);current_path.pop_back();}}}bool compare(vector<int> p1, vector<int> p2){int minlen=(p1.size()<p2.size())?p1.size():p2.size();for(int i=1;i<minlen;i++){if(tree[p1[i]].weight>tree[p2[i]].weight)return true;else if(tree[p1[i]].weight<tree[p2[i]].weight)return false;elsecontinue;}return false;}int main(){int n_nodes,n_non_leaf,given_s;cin>>n_nodes>>n_non_leaf>>given_s;for(int i=0;i<n_nodes;i++)cin>>tree[i].weight;for(int i=0;i<n_non_leaf;i++){int id,k;cin>>id>>k;for(int j=0;j<k;j++){int node;cin>>node;tree[id].child.push_back(node);}}current_path.push_back(0);FindAllPath(given_s,0);//first we need to sort the resultsort(path.begin(),path.end(),compare);for(int i=0;i<(int)path.size();i++){vector<int> p = path[i];for(int j=0;j<(int)p.size();j++){if(j==(int)p.size()-1)cout<<tree[p[j]].weight<<endl;elsecout<<tree[p[j]].weight<<" ";}}return 0;}