PAT1053. 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_i for i=1, ... k, and A_k+1 > B_k+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
思路:DFS  建立好树

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <vector>
 4 #include <algorithm>
 5 using namespace std;
 6 #define MAX 110
 7 struct Node
 8 {
 9     int weight;
10     vector<int>child;
11 }node[MAX];
12 int path[MAX];
13 int pt=0;
14 int N,M,S;
15 bool cmp(int a,int b)
16 {
17     return node[a].weight>node[b].weight;
18 }
19 void Print()
20 {
21     for(int i=0;i<pt;i++)
22     {
23         printf("%d",node[path[i]].weight);
24         if(i!=pt-1)
25           putchar(‘ ‘);
26     }
27     putchar(‘\n‘);
28 }
29 int sum=0;
30 void DFS(int index)
31 {
32     sum+=node[index].weight;
33     path[pt++]=index;
34     if(sum>=S||node[index].child.size()==0)
35     {
36         if(sum==S&&node[index].child.size()==0)
37            Print();
38         sum-=node[index].weight;
39         pt--;
40         return;
41     }
42     for(int i=0;i<node[index].child.size();i++)
43     {
44         DFS(node[index].child[i]);
45     }
46     sum-=node[index].weight;
47     pt--;
48 }
49 int main(int argc, char *argv[])
50 {
51      scanf("%d%d%d",&N,&M,&S);
52      for(int i=0;i<N;i++)
53      {
54          scanf("%d",&node[i].weight);
55      }
56      for(int i=0;i<M;i++)
57      {
58          int father,k;
59          scanf("%d%d",&father,&k);
60          for(int i=0;i<k;i++)
61          {
62              int tem;
63              scanf("%d",&tem);
64              node[father].child.push_back(tem);
65         }
66         sort(node[father].child.begin(),node[father].child.end(),cmp);
67       }
68       DFS(0);
69     return 0;
70 }