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<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

Special thanks to Zhang Yuan and Yang Han for their contribution to the judge's data.

---

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
struct Tnode {
	int val;
	vector<int> ch;
}tnode[110];
int n, m, s;
vector<int> ans[110], t;
int cnt = 0, u = 0;
void dfs(int root) {
	cnt += tnode[root].val;
	t.push_back(tnode[root].val);
	if (cnt < s) {
		for (int i = 0; i < tnode[root].ch.size(); i++) {
			dfs(tnode[root].ch[i]);
		}
	}
	else if (cnt > s) {
		cnt -= tnode[root].val;
		t.pop_back();
		return;
	}
	else {
		if (tnode[root].ch.size() == 0) {
			ans[u] = t;
			u++;
		}
	}
	cnt -= tnode[root].val;
	t.pop_back();
}
bool cmp(vector<int> a, vector<int> b) {
	return a > b;
}
int main() {
	cin >> n >> m >> s;
	for (int i = 0; i < n; i++) {
		scanf("%d", &tnode[i].val);
	}
	for (int i = 0; i < m; i++) {
		int idx, num;
		scanf("%d%d", &idx, &num);
		for (int j = 0; j < num; j++) {
			int tmp;
			scanf("%d", &tmp);
			tnode[idx].ch.push_back(tmp);
		}
	}
	dfs(0);
	sort(ans, ans + u, cmp);
	for (int i = 0; i < u; i++) {
		for (int j = 0; j < ans[i].size(); j++) {
			if (j == 0) printf("%d", ans[i][j]);
			else printf(" %d", ans[i][j]);
		}
		printf("\n");
	}
	return 0;
}