1053. Path of Equal Weight (30)

dfs函数携带vector形参记录搜索路径

时间限制

10 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 = Bifor i=1, ... k, and Ak+1 > Bk+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

来源: <http://www.patest.cn/contests/pat-a-practise/1053>

  1. #include <iostream>
  2. #include <map>
  3. #include <vector>
  4. #include <algorithm>
  5. using namespace std;
  6. map<int, vector<int>> mp;
  7. vector<int> weight;
  8. int n, m, aimWeight;
  9. int pathWeight=0;
  10. bool cmp(int a, int b) {
  11. return weight[a] > weight[b];
  12. }
  13. void dfs(int i,int w, vector<int> p) {
  14. p.push_back(i);
  15. if (mp[i].size() == 0) {
  16. if (w == aimWeight) {
  17. for (int j = 0; j < p.size(); j++) {
  18. cout << weight[p[j]];
  19. if (j != p.size() - 1)
  20. cout << " ";
  21. else
  22. cout << endl;
  23. }
  24. }
  25. p.clear();
  26. return;
  27. }
  28. else {
  29. for (int j = 0; j < mp[i].size(); j++)
  30. dfs(mp[i][j],weight[mp[i][j]]+w,p);
  31. }
  32. }
  33. int main(void) {
  34. cin >> n >> m >> aimWeight;
  35. for (int i = 0; i < n; i++) {
  36. int wtemp;
  37. cin >> wtemp;
  38. weight.push_back(wtemp);
  39. }
  40. for (int i = 0; i < m; i++) {
  41. int id, k;
  42. cin >> id >> k;
  43. for (int j = 0; j < k; j++) {
  44. int idtemp;
  45. cin >> idtemp;
  46. mp[id].push_back(idtemp);
  47. }
  48. sort(mp[id].begin(), mp[id].end(), cmp);
  49. }
  50. vector<int> path;
  51. dfs(0,weight[0],path);
  52. return 0;
  53. }

来自为知笔记(Wiz)

时间: 12-03

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