1053 Path of Equal Weight （30 分)

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

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≤kB​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

题意:给出一棵树和每个结点权值，求权值之和为s的从根到叶的结点权值路径。并且这条路径按权值大小非递增排列(是不同路径的权值比较)。

思路:1.可以用结构体存储树的权值和子树。因为得出的结果要按权值大小排序，所以读入时先对每个结点排序，这样遍历时就会优先遍历权值大的结点了。

2.用path数组存储路径，即结点的编号。和dijkstra不同,不需要temppath数组，因为这条路径遍历时是不会改变的。然后用dfs遍历树找路径。index为当前结点，numNode为结点个数（也是递归层数，每深入一层，path就会多一个结点），sum为当前路径的权值。

注意:这里的排序是在遍历之前就进行的。

#include 
#include 
#include 
using namespace std;

const int maxn=110;
int n,m,s;
int path[maxn];
struct node{
	int weight;//权值 
	vector child;//子树 
}Node[maxn];

bool cmp(int a,int b)
{
	return Node[a].weight>Node[b].weight;
}

void dfs(int index,int numNode,int sum)
{
	if(sum>s){
		return ;
	}
	else if(sum==s){
		if(Node[index].child.size()!=0){//如果此时权值相等还有子树，那么到叶节点时权值一定会大于s 
			return ;
		}
		for(int i=0;i