1053. Path of Equal Weight (30)
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1053. Path of Equal Weight (30)
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.
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 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 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 19Sample Output:
10 5 2 710 4 1010 3 3 6 210 3 3 6 2
树的结构可以用邻接表存储,然后使用dfs很容易可以找到满足要求的一些路径,然后就是对这些路径按要求进行排序,使用sort很容易就可以完成。
#include <iostream>#include <vector>#include <map>#include <queue>#include <algorithm>using namespace std;void dfs(int root,int total_weight);bool compare(vector<int> a,vector<int> b);vector<int> weight;map<int,vector<int> > tree;vector<vector<int> > record;vector<int> weight_record;int target,cnt=0;int main(){ int num_node,num_none_leaf_node; cin >>num_node >>num_none_leaf_node >>target; for(int i=0;i<num_node;i++) { int tmp; cin >>tmp; weight.push_back(tmp); } for(int i=0;i<num_none_leaf_node;i++) { int root,num; cin >>root >>num; for(int j=0;j<num;j++) { int tmp; cin >>tmp; tree[root].push_back(tmp); } } weight_record.push_back(weight[0]); dfs(0,weight[0]); sort(record.begin(),record.end(),compare); int max_step=record.size(); for(int i=0;i<max_step;i++) { vector<int> tmp=record[i]; cout <<tmp[0]; for(int j=1;j<tmp.size();j++) cout <<" "<<tmp[j]; cout <<endl; } return 0;}void dfs(int root,int total_weight){ //到达一个叶子节点 if(tree.find(root)==tree.end()) { if(total_weight==target) { record.push_back(weight_record); cnt++; } return; } //没有到达叶子节点,但是权重已经大于target,终止dfs if(total_weight>=target) return; //没有到达叶子节点,但是权重仍然小于target int num_child=tree[root].size(); for(int i=0;i<num_child;i++) { int tmp=tree[root].at(i); int a=total_weight+weight[tmp]; weight_record.push_back(weight[tmp]); dfs(tmp,a); weight_record.pop_back(); }}//降序排列bool compare(vector<int> a,vector<int> b){ int length=min(a.size(),b.size()); for(int i=0;i<length;i++) if(a[i]!=b[i]) return a[i]>b[i]; return a.size()>b.size();}
- 1053. Path of Equal Weight (30)
- 1053. Path of Equal Weight (30)-PAT
- 1053. Path of Equal Weight (30)
- 1053. Path of Equal Weight (30)
- PAT 1053. Path of Equal Weight (30)
- 1053. Path of Equal Weight (30)
- 1053. Path of Equal Weight (30)
- 1053. Path of Equal Weight (30)
- 1053. Path of Equal Weight (30)
- PAT 1053. Path of Equal Weight (30)
- 1053. Path of Equal Weight (30)
- 1053. Path of Equal Weight (30)
- 1053. Path of Equal Weight (30)
- 1053. Path of Equal Weight (30)
- PAT 1053. Path of Equal Weight (30)
- 1053. Path of Equal Weight (30)
- 1053. Path of Equal Weight (30)
- 1053. Path of Equal Weight (30)
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