1053. Path of Equal Weight (30)

Given a non-empty tree with root R, and with weight W_i assigned to each tree node T_i. Theweight 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 begreater than sequence {B₁, B₂, ..., B_m} if there exists 1 <= k < min{n, m} such that A_i = B_i for 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

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在一棵树中找出节点的值总和为m的路径，输出按照指定的顺序。先按照输入的节点信息建立一棵树，然后对这棵树进行深度优先搜索，找出节点数值的总和为m的路径，把路径储存在一个vector数组中，然后用sort函数进行排序（比较函数自己写），最后输出即可。

代码：

#include <iostream>#include <cstring>#include <vector>#include <cstdlib>#include <cstdio>#include <algorithm>using namespace std;struct node{int id;int val;bool isleaf;vector<int>next;node():isleaf(true){}};vector<node>tree;vector<vector<int> >res;void dfs(int cur,int sum,int target,vector<int>path){sum+=tree[cur].val;path.push_back(tree[cur].val);if(tree[cur].isleaf){if(sum==target){res.push_back(path);}return;}int n=tree[cur].next.size();for(int i=0;i<n;i++){dfs(tree[cur].next[i],sum,target,path);}}bool cmp(vector<int>v1,vector<int>v2){for(int i=0;i<v1.size()&&i<v2.size();i++){if(v1[i]!=v2[i]) return v1[i]>v2[i];}return false;}int main(){int n,m,s;cin>>n>>m>>s;tree.resize(n);for(int i=0;i<n;i++){cin>>tree[i].val;}for(int i=0;i<m;i++){int id,k;cin>>id>>k;tree[id].isleaf=false;for(int j=0;j<k;j++){int next;cin>>next;tree[id].next.push_back(next);}}vector<int>path;dfs(0,0,s,path);sort(res.begin(),res.end(),cmp);for(int i=0;i<res.size();i++){cout<<res[i][0];for(int j=1;j<res[i].size();j++){cout<<" "<<res[i][j];}cout<<endl;}}