PAT 1053 Path of Equal Weight [DFS] [树的遍历]

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<2​30​​, 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 {A​1​​,A​2​​,⋯,A​n​​} is said to be greater than sequence {B​1​​,B​2​​,⋯,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

----------------------------------这是题目和解题的分割线----------------------------------

输出权重和为给定值的路径，用DFS。静态存储这棵树，便于操作。

#include<cstdio>
#include<vector>
#include<algorithm>

using namespace std;

struct node
{
	int weight;
	//子结点是不确定的，为了避免可能存在的内存超限，用可变数组vector 
	vector<int> child;  
}node[110];

int m,n,s;
vector<int> tmp;

void dfs(int index,int sum)
{
	if(sum>s) return; //大于，中止
	//找到，并且保证这是叶子结点，不然第0个测试点会答案错误 
	if(sum==s&&node[index].child.size()==0)
	{
		for(int i=0;i<tmp.size();i++)
		{
			printf("%d",node[tmp[i]].weight); //输出weight 
			if(i!=tmp.size()-1) printf(" ");
		}
		printf("\n");
	}
	for(int i=0;i<node[index].child.size();i++)
	{ 
		int ch = node[index].child[i];
		tmp.push_back(ch); //记录下这个结点 
		dfs(ch,sum+node[ch].weight); //递归 
		tmp.pop_back(); //回溯后pop，以免影响下一次递归 
	}
}

//weight从大到小排序 
bool cmp(int a,int b)
{
	return node[a].weight>node[b].weight;
}

int main()
{
	int i,j,id,x,child;
	scanf("%d%d%d",&n,&m,&s);
	for(i=0;i<n;i++)
		scanf("%d",&node[i].weight);
	for(i=0;i<m;i++)
	{
		scanf("%d%d",&id,&x);
		for(j=0;j<x;j++)
		{
			scanf("%d",&child); //读入结点关系 
			node[id].child.push_back(child);
		}
		//子结点排序成从大到小的顺序，便于最后输出 
		//begin end的用法  
		sort(node[id].child.begin(),node[id].child.end(),cmp);
	}
	//先push头结点，不然等会儿不太好操作 
	tmp.push_back(0);
	dfs(0,node[0].weight); //push头结点要把头结点的weight也传进去 
	return 0;
}