• Path of Equal Weight (DFS)


    Path of Equal Weight (DFS)

     

    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 = Bi for 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

     

     

    这道30分的题目,提交一次就意外的AC了。

     

    就是 建立连接表 DFS+记录路径+权值累加 搜到叶子节点,如果权值之和与要求的的相等时保存路径。

     

    最后的排序要点混,但进行三层的判断排序,也就能过了,


    #include <iostream>
    
    #include <string>
    
    #include <vector>
    
    #include <algorithm>
    
    using namespace std;
    
     
    
    int WW[100];
    
    int visit[100];
    
    vector<int> vv[100];
    
     
    
    vector<int> road;
    
     
    
    vector<int> RR[100];
    
     
    
    int sum,wi;
    
     
    
    bool cmp(vector<int> a,vector<int> b)
    
    {
    
          if(a[0]==b[0]&&a[1]==b[1])
    
                return a[2]>b[2];
    
          if(a[0]==b[0])
    
          return      a[1]>b[1];
    
     
    
          return      a[0]>b[0];
    
     
    
    }
    
     
    
    void DFS(int root,int &count)
    
    {
    
        if(visit[root]==0)
    
          {
    
             visit[root]=1;
    
             road.push_back(WW[root]);
    
           sum+=WW[root];
    
     
    
             for(int i=0;i<vv[root].size();i++)
    
             {
    
                if(visit[vv[root][i]]==0)
    
                        DFS(vv[root][i],count);
    
             }
    
             if(sum==wi&&vv[root].size()==0)
    
             {
    
                RR[count++]=road;
    
                
    
             }
    
              road.pop_back();
    
                sum-=WW[root];
    
     
    
          }
    
    }
    
     
    
     
    
     
    
    int main()
    
    {
    
         
    
          int i,j,num,fnum;
    
          while(cin>>num)
    
          {
    
               
    
          road.clear(); 
    
     
    
            cin>>fnum>>wi;
    
            for(i=0;i<num;i++)
    
            {
    
                  cin>>WW[i];
    
                  vv[i].clear();
    
                  visit[i]=0;
    
                  RR[i].clear();
    
            }
    
     
    
            for(i=0;i<fnum;i++)
    
            {
    
                  int n1,n2;
    
               cin>>n1>>n2;
    
                for(j=0;j<n2;j++)
    
                {
    
                    int tem;
    
                      cin>>tem;
    
                      vv[n1].push_back(tem);
    
                }
    
            }
    
     
    
          int count=0;
    
             sum=0;
    
             DFS(0,count);
    
     
    
          sort(RR,RR+count,cmp);
    
          
    
            for(i=0;i<count;i++)
    
            {
    
               cout<<RR[i][0];
    
                for(j=1;j<RR[i].size();j++)
    
                      cout<<" "<<RR[i][j];
    
                cout<<endl;
    
            }
    
          }
    
     
    
      return 0;
    
    }


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  • 原文地址:https://www.cnblogs.com/xiaoyesoso/p/4234436.html
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