【PAT甲级】1053 Path of Equal Weight (30分)：DFS（树的先序遍历）

题目：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,⋯,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

      
    

分析：DFS（树的先序遍历）

题意：输出点权和为S的路径，以字典序。

• DFS遍历时计算一条路径的weight，只需要在每次递归DFS时传参weight+Node[index].data即可。

不要单独拿出来一句语句weight+=Node[index].data，因为和临时路径temp一样（每次push新节点后还要pop以保证其他路径没有收到影响），相应语句weight-=Node[index].data 很容易忘记写导致错误。

• 最后路径需要按照字典序输出，有两种方法：
  • 方一：设置变量vector<vector<int>> ans，将所有路径加到ans，最后用sort（）对ans元素进行排序。
  • 方二：在未进行DFS前，用sort()先对输入的树的数据进行结构调整，使路径一开始按照字典序排列的，之后进行DFS遍历时就可以直接输出路径了。

代码

#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;
}

注意

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

  ”->“前不是指针。

对象调用成员变量用.

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

//Node[root].data

指针引用成员变量->

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

//Node->rchild

• 用sort()对vector<vector<int>>排序
  • sort()参数写ans.begin()和ans.end()。
  • 注意比较器的写法

#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;
}