PAT1053. Path of Equal Weight

Given a non-empty tree with the root R, and with the weight Wi assigned to each tree node Ti. The weight of a path from R to L are defined to being the sum of the weights of the nodes along the path from R t o Any leaf node L.

Now given any weighted tree, you is 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 are the node ID which is a two-dig  It number, and the lower number is the weight of that node. Suppose the given number is and then there exists 4 different paths which has the same given weight: {10 5 2 7}, {10 4}, {3 3 6 2} and {ten 3 3 6 2}, which correspond to the red edges in Figure 1.

Figure 1

Input Specification:

Each input file contains the one test case. Each case starts with a line containing 0 < N <=, the number of nodes in a tree, M (< N), the number of non-l EAF nodes, and 0 < S <, 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 was 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 is separated by a space with no extra space at the end of the line.

note:sequence {A1, A2, ..., an} are 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 19

Sample Output:

10 5 2 710 4 1010 3 3 6 210 3 3 6 2
Idea: DFS  builds a good tree

1#include <iostream>2#include <cstdio>3#include <vector>4#include <algorithm>5 using namespacestd;6 #defineMAX 1107 structNode8 {9     intweight;Tenvector<int>Child ;  One }node[max]; A intPath[max]; - intpt=0; - intn,m,s; the BOOLcmpintAintb) - { -     returnNode[a].weight>Node[b].weight; - } + voidPrint () - { +      for(intI=0; i<pt;i++) A     { atprintf"%d", node[path[i]].weight); -         if(i!=pt-1) -Putchar (' '); -     } -Putchar ('\ n'); - } in intsum=0; - voidDFS (intindex) to { +sum+=Node[index].weight; -path[pt++]=index; the     if(sum>=s| | Node[index].child.size () = =0) *     { $         if(sum==s&&node[index].child.size () = =0)Panax Notoginseng Print (); -sum-=Node[index].weight; thept--; +         return; A     }     the      for(intI=0; I<node[index].child.size (); i++) +     { - DFS (Node[index].child[i]); $     } $sum-=Node[index].weight; -pt--; - } the intMainintargcChar*argv[]) - {Wuyiscanf"%d%d%d",&n,&m,&S); the       for(intI=0; i<n;i++) -      { Wuscanf"%d",&node[i].weight); -      } About       for(intI=0; i<m;i++) $      { -          intfather,k; -scanf"%d%d",&father,&k); -           for(intI=0; i<k;i++) A          { +              inttem; thescanf"%d",&tem); - Node[father].child.push_back (TEM); $         } the sort (Node[father].child.begin (), Node[father].child.end (), CMP); the       } theDFS (0); the     return 0; -}

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PAT1053. Path of Equal Weight