PAT A1053：Path of Equal Weight

题目描述

1053 Path of Equal Weight (30分)
Given a non-empty tree with root R, and with weight $W_i$Wi​ assigned to each tree node $T_i$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 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<2^30$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 ${A_1 ,A_2 ,……,A_n }$A1​,A2​,……,An​ is said to be greater than sequence ${B_1 ,B_2 ,……,B_m }$B1​,B2​,……,Bm​ if there exists $1≤k<min\left(n,m\right)$1≤k<min(n,m) such that $A_i =B_i$Ai​=Bi​ for i=1,…，k, and $A_{k+1} >B_{k+1}$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

求解思路

给定一棵树和每个结点的权值，求所有从根节点到叶子结点的路径，使得每条路径上的结点的权值之和等于给定的常数S。如果有多条这样的路径，则按路径非递增的顺序输出。

• 采用树结构，定义一个结构体node，用于存放结点的数据，以及指向其子节点的指针，因为子节点可能不止一个，所用用一个vector来存储每个结点的编号。因为最后的输出需要按照权值从大到小排序，所以在读入的时候可以实现对每个结点的子节点进行排序。这样就可以优先遍历每个子节点中权值更大的那个子节点
• 用一个vector容器path来保存递归过程中产生的路径上的结点编号。接下来进行DFS搜索，参数：当前访问的结点标号index、当前路径上的权值和sum。递归过程的伪代码如下：
  • 若sum>S，直接return
  • 若sum==S，说明到达当前访问结点index为止，输入中需要达到的S已经得到了。如果当前结点位叶子结点，则输出path中的所有数据，否则return
  • 若sum<S，说明要求还没满足。此时枚举当前访问结点index的所有子节点，对每一个子节点child，先将其存入path中，然后在此基础上往下一层递归，下一层递归的参数为child，sum+node[child].weight
  • 注意：使用vector存储路径的话，记得下一层回溯上来之后将前面加入的子节点pop_back出来

代码实现

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn=10001;
struct Node{
	int weight;
	vector<int>child;
}node[maxn];
int n,m,s;
vector<int>path;
bool cmp(int a,int b)	//根据结点值对结点进行排序 
{
	return node[a].weight>node[b].weight;
}
void DFS(int index,int sum)
{
	if(sum>s)	return;
	else if(sum==s)
	{
		if(node[index].child.size())	return;	//非叶子结点
		else
		{
			for(int i=0;i<path.size();i++)
			{
				printf("%d",node[path[i]].weight);
				if(i<path.size()-1)	printf(" ");
				else printf("\n");
			}	
		}	
	}
	else
	{
		for(int i=0;i<node[index].child.size();i++)	
		{
			int child=node[index].child[i];
			path.push_back(child);
			DFS(child,sum+node[child].weight);
			path.pop_back();	//下一层递归回溯上来之后将前面加入的子节点pop_back即可 
		}
	}
}
void init()
{
	// 用于数据读取与初始化 
	scanf("%d %d %d",&n,&m,&s);
	for(int i=0;i<n;i++)
	{
		scanf("%d",&node[i].weight);
	}
	for(int i=0;i<m;i++)
	{
		int node_index,child_num;
		scanf("%d %d",&node_index,&child_num);
		for(int j=0;j<child_num;j++)
		{
			int child_index;
			scanf("%d",&child_index);
			node[node_index].child.push_back(child_index);
		}
		sort(node[node_index].child.begin(),node[node_index].child.end(),cmp);
	}	
}
void solve()
{
	init();
	path.push_back(0);	//路径第一个结点设置为0号结点 
	DFS(0,node[0].weight);
}
int main()
{
	solve();
	return 0;
}