PAT A1053. Path of Equal Weight (30) [树的遍历]

Given a non-empty tree with root R, and with weight Wi assigned to each tree node 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 Figure 1: 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 diferent 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 Figure 1.

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 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 {A1, A2, …, An} is said to be greater than sequence {B1, B2, …, Bm} if there exists 1 <= k <
min{n, m} such that Ai = Bi for i=1, … k, and Ak+1 > Bk+1
Sample Input:
20 9 24总结点数——非叶结点数——待判定权值
10 2 4 3 5 10 2 18 9 7 2 2 1 3 12 1 8 6 2 2按结点顺序Ti给出权值Wi
00 4 01 02 03 04结点ID——孩子数——孩子ID
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
题⽬⼤意：给出树的结构和权值，找从根结点到叶⼦结点的路径上的权值相加之和等于给定⽬标数的
路径，并且从⼤到⼩输出路径
分析：对于接收孩⼦结点的数据时，每次完全接收完就对孩⼦结点按照权值进⾏排序（序号变，根据
权值变），这样保证深度优先遍历的时候直接输出就能输出从⼤到⼩的顺序。记录路径采取这样的⽅
式：⾸先建⽴⼀个path数组，传⼊⼀个nodeNum记录对当前路径来说这是第⼏个结点（这样直接在
path[nodeNum]⾥⾯存储当前结点的孩⼦结点的序号，这样可以保证在先判断return的时候， path是从
0~numNum-1的值确实是要求的路径结点）。然后每次要遍历下⼀个孩⼦结点的之前，令
path[nodeNum] = 孩⼦结点的序号，这样就保证了在return的时候当前path⾥⾯从0~nodeNum-1的值就
是要输出的路径的结点序号，输出这个序号的权值即可，直接在return语句⾥⾯输出。
注意：当sum==target的时候，记得判断是否孩⼦结点是空，要不然如果不空说明没有到底部，就直接
return⽽不是输出路径。。。（这对接下来的孩⼦结点都是正数才有⽤，负权⽆⽤）。

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int target;
struct NODE {
	int w;
	vector<int> child;
	};
vector<NODE> v;
vector<int> path;
void dfs(int index, int nodeNum, int sum) {
	if(sum > target) return ;
	if(sum == target) {
		if(v[index].child.size() != 0) return;
		for(int i = 0; i < nodeNum; i++)
			printf("%d%c", v[path[i]].w, i != nodeNum - 1 ? ' ' : '\n');
		return ;
	}
	for(int i = 0; i < v[index].child.size(); i++) {
		int node = v[index].child[i];
		path[nodeNum] = node;
		dfs(node, nodeNum + 1, sum + v[node].w);
	}
}
int cmp1(int a, int b) {
	return v[a].w > v[b].w;
}
int main() {
	int n, m, node, k;
	scanf("%d %d %d", &n, &m, &target);
	v.resize(n), path.resize(n);
	for(int i = 0; i < n; i++)
		scanf("%d", &v[i].w);
	for(int i = 0; i < m; i++) {
		scanf("%d %d", &node, &k);
		v[node].child.resize(k);
		for(int j = 0; j < k; j++)
		scanf("%d", &v[node].child[j]);
		sort(v[node].child.begin(), v[node].child.end(), cmp1);
	}
	dfs(0, 1, v[0].w);
return 0;
}