PAT A1053 Path of Equal Weight（30 分）

时间限制: 400 ms

内存限制: 64 MB

代码长度限制: 16 KB

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<, 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 2410 2 4 3 5 10 2 18 9 7 2 2 1 3 12 1 8 6 2 200 4 01 02 03 0402 1 0504 2 06 0703 3 11 12 1306 1 0907 2 08 1016 1 1513 3 14 16 1717 2 18 19

Sample Output:

10 5 2 7

10 4 10

10 3 3 6 2

10 3 3 6 2
代码如下：

package patA;

import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.Scanner;
import java.util.Vector;

public class A1053 {

	static int path[] = new int[100];// 记录路径
	static int number;// 记录总结点数

	static int notleaf;// 非叶子结点数

	static int mubiao;// 给定的值

	static int quanzhi[] = new int[100];
	// 结点数组
	static Node node[] = new Node[100];

	// 结点的类
	static class Node {
		int data;// 每个点的权值

		Vector<Integer> v = new Vector();// 存储孩子结点
	}

	// 深度遍历算法,index是结点编号，numNode是path路径上的结点个数，sum是path 上的结点和
	static void dfs(int index, int numNode, int sum) {
		
	
// 如果到跟结点前已经超过目标
		if(sum>mubiao) {
			
		
			return ;
		}else if(sum==mubiao) {
			
			
			
			
			if(node[index].v.size()!=0) {
				return ;
			}else{
				//到达了叶子结点，输出path
				for(int m=0;m<numNode;m++) {
					System.out.print(node[path[m]].data);
				}
			}
		}
		
		for(int h=0;h<node[index].v.size();h++) {
			int child=node[index].v.get(h);
			path[numNode]=child;
			dfs(child,numNode+1,sum+node[child].data);
		}
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);

		number = sc.nextInt();
		notleaf = sc.nextInt();
		mubiao = sc.nextInt();

		for (int i = 0; i < number; i++) {
			quanzhi[i] = sc.nextInt();
			node[i] = new Node();
			node[i].data=quanzhi[i];
		}
		// 初始化Node数组
		for (int i = 0; i < notleaf; i++) {
			int num = sc.nextInt();

			int numchild = sc.nextInt();
			for (int j = 0; j < numchild; j++) {
	
				node[num].v.add(sc.nextInt());

				// 对子节点按照从大到小排序
				Compaa compaa = new Compaa();
				Collections.sort(node[num].v, compaa);
			}

		}
		
		path[0]=0;
		dfs(0,1,node[0].data);

	}

}

class Compaa implements Comparator<Integer> {
	public int compare(Integer n1, Integer n2) {
		if (n1 > n2) {
			return -1;
		} else {
			return 1;
		}
	}
}