PAT 1053 Path of Equal Weight (30)

1053 Path of Equal Weight (30)（30 分）

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 < 2^30^, the given weight number. The next line contains N positive numbers where W~i~ (&lt1000) 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 {A~1~, A~2~, ..., A~n~} is said to be greater than sequence {B~1~, B~2~, ..., 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

 1 #include<iostream>
 2 #include<vector>
 3 #include<algorithm>
 4 using namespace std;
 5 vector<int> ans[100], v[100];
 6 vector<int> weight, temp;
 7 int s, cnt=0;
 8 
 9 void dfs(int root, int w){
10   temp.push_back(weight[root]);
11   if(v[root].size()==0){
12     if(s==w) ans[cnt++]=temp;
13     temp.pop_back();
14     return;
15   }
16   for(int i=0; i<v[root].size(); i++)
17     if(w+weight[v[root][i]]<=s)
18       dfs(v[root][i], w+weight[v[root][i]]);
19   temp.pop_back();
20 }
21 
22 int main(){
23   int n, m, i;
24   cin>>n>>m>>s;
25   weight.resize(n);
26   for(i=0; i<n; i++) scanf("%d", &weight[i]);
27   int k, id, child, j;
28   for(i=0; i<m; i++){
29     scanf("%d %d", &id, &k);
30     for(j=0; j<k; j++){
31       scanf("%d", &child);
32       v[id].push_back(child);
33     }
34   }
35   dfs(0, weight[0]);
36   sort(ans, ans+cnt);
37   
38   for(i=cnt-1; i>=0;i--) {
39     for(int j=0; j<ans[i].size(); j++){
40       if(j==0) cout<<ans[i][j];
41       else cout<<" "<<ans[i][j];
42     }
43     cout<<endl;
44   }
45   return 0;
46 }