1053. Path of Equal Weight (30)

题目连接：https://www.patest.cn/contests/pat-a-practise/1053原题如下：

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

这道是我觉得是一道很典型的DFS题。我自己递归水平不行，参考了陈小旭的解法。值得多次回看！！！

 1 #include<stdio.h>
 2 #include<vector>
 3 #include<algorithm>
 4 #define MAXN 105
 5 using namespace std;
 6 
 7 typedef struct Node
 8 {
 9     int weight;
10     vector<int>son;
11 }node;
12 node bur[MAXN];
13 int S;
14 vector<int>ans;
15 vector<vector<int> >ans_V;
16 
17 void DFS(int p,int sum)
18 {
19     sum+=bur[p].weight;
20     ans.push_back(bur[p].weight);
21     if (sum<S)
22     {
23         for (int i=0;i<bur[p].son.size();i++)DFS(bur[p].son[i],sum);
24         ans.pop_back();  //没有孩子，但仍然小于S
25         return ;
26     }
27     else if (sum>S)
28     {
29         ans.pop_back();
30         return ;
31     }
32     else if (sum==S)
33     {
34         if (bur[p].son.size()==0)
35         {
36             ans_V.push_back(ans);
37             ans.pop_back();
38         }
39         else
40         {
41             ans.pop_back();
42         }
43         return ;
44     }
45 }
46 
47 int cmp(vector<int>a,vector<int>b)
48 {
49     int len1,len2;
50     len1=a.size();
51     len2=b.size();
52 
53     int i;
54     for (i=0;i<len1 && i<len2;i++)
55     {
56         if (a[i]==b[i])continue;
57         else  return (a[i]>b[i]);
58     }
59     return (len1>len2);
60 }
61 
62 int main()
63 {
64     int N,M;
65     scanf("%d %d %d",&N,&M,&S);
66     for (int i=0;i<N;i++)scanf("%d",&bur[i].weight);
67     int p,k,w;
68     while(M--)
69     {
70         scanf("%d %d",&p,&k);
71         while(k--)
72         {
73             scanf("%d",&w);
74             bur[p].son.push_back(w);
75         }
76     }
77     DFS(0,0);
78     sort(ans_V.begin(),ans_V.end(),cmp);
79     for (int i=0;i<ans_V.size();i++)
80     {
81         for (int j=0;j<ans_V[i].size();j++)
82         {
83             if (j==0)printf("%d",ans_V[i][j]);
84             else printf(" %d",ans_V[i][j]);
85         }
86         printf("
");
87     }
88     return 0;
89 }