[PAT] Grade 1053 Path of Equal Weight (30 points): DFS (preorder traversal of the tree)

Topic: https://pintia.cn/problem-sets/994805342720868352/problems/994805424153280512

1053 Path of Equal Weight (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 the following figure: 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 the figure.

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,⋯,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

      
    

Analysis: (preorder traversal of the tree) DFS

Meaning of the title: the right point and the output path S to the dictionary order.

• Calculating a path traversing the DFS weight, need only pass parameters at each recursive DFS weight+Node[index].datacan.

Do not separate out a statement weight+=Node[index].data, because the same temp and temporary path (but also pop each other to ensure that the path does not affect the push receive new node), the corresponding statement weight-=Node[index].datais easy to forget to write causes an error.

• Finally, the output path requires lexicographic order, there are two methods:
  • A side: variable settings vector<vector<int>> ans, all paths ans was added, and finally sort () to sort the elements ans.
  • Side two: not performed before DFS, with sort () to the data input tree structure is adjusted so that the beginning of the path are arranged in lexicographic order, performed after the DFS traversal path can be output directly.

Code

#include<iostream>
#include<vector>
#include<algorithm>

using namespace std;

const int maxN=110;
struct node {
    int data=0;
    vector<int> child;
} Node[maxN];
int N,M,S;

vector<vector<int>> ans;//路径集合
vector<int> temp;//临时路径

void DFS(int root,int weight) {
    if(weight>S) {
        return;
    }
    if(weight==S) {
        if(Node[root].child.size()==0) {
            ans.push_back(temp);
        }
        return;
    }

    for(int i=0; i<Node[root].child.size(); i++) {
        int index=Node[root].child[i];
        temp.push_back(Node[index].data);
        DFS(index,weight+Node[index].data);
        temp.pop_back();
    }

}

int main() {
    scanf("%d%d%d",&N,&M,&S);
    for(int i=0; i<N; i++) {
        scanf("%d",&Node[i].data);
    }
    int index,childNum,childNo;
    for(int i=0; i<M; i++) {
        scanf("%d%d",&index,&childNum);
        for(int j=0; j<childNum; j++) {
            scanf("%d",&childNo);
            Node[index].child.push_back(childNo);
        }
    }

    temp.push_back(Node[0].data);
    DFS(0,Node[0].data);
    
    sort(ans.begin(),ans.end(),greater<vector<int>>());//从大到小排序ans
    for(int i=0; i<ans.size(); i++) {
        for(int j=0; j<ans[i].size(); j++) {
            printf("%d",ans[i][j]);
            if(j<ans[i].size()-1)
                printf(" ");
            else
                printf("\n");
        }
    }

    return 0;
}

note

• [Error] base operand of '->' has non-pointer type 'node'

  "->" is not a front pointer.

Object calls member variables.

struct node {
    int data=0;
    vector<int> child;
} Node[maxN];

//Node[root].data

Pointer reference member variables->

struct node {
    int data;
    node* lchild;
    node* rchild;
}Node;

//Node->rchild

• () Sorted vector <vector <int >> sort by
  • sort () parameter write ans.begin () and ans.end ().
  • Note comparator wording

#include<iostream>
#include<vector>
#include<algorithm>

using namespace std;

vector<int> v1= {10,5,2,7};
vector<int> v2= {10,3,3,6,2};
vector<int> v3= {10,3,3,6,2};
vector<int> v4= {10,4,10};
vector<vector<int>> ans;

bool cmp(vector<int> v1,vector<int> v2) {
    return v1>v2;
}

//比较器的错误写法 
//bool cmp(vector<int> v1,vector<int> v2) {
//  if(v1>v2) return v1>v2;
//  else return v1<v2;
//}

int main() {
    ans.push_back(v1);
    ans.push_back(v2);
    ans.push_back(v3);
    ans.push_back(v4);
//  sort(ans[0],ans[0]+2*2,cmp);//错误：取到了一个vector，是对象
    sort(ans.begin(),ans.end(),cmp);//取到了一个迭代器//方一
//  sort(ans.begin(),ans.end(),greater<vector<int>>());//方二 
    for(int i=0; i<ans.size(); i++) {
        vector<int> temp=ans[i];
        for(auto x:temp)
            cout<<x<<" ";
        cout<<endl;

    }


    return 0;
}