1053. Path of Equal Weight (30)

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本文介绍了一个算法问题，即在一个加权树中查找所有路径，这些路径的权重之和等于给定数值。通过深度优先搜索（DFS）算法实现，并对输入输出格式进行了详细说明。

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 Lis 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 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 Figure 1.

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 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 {A₁, A₂, ..., A_n} is said to be greater than sequence {B₁, B₂, ..., B_m} if there exists 1 <= k < min{n, m} such that A_i = B_ifor 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:

// dfs题。
#include "iostream"
#include "vector"
#include "algorithm"
using namespace std;
int n, m, s;
int w[102];
int ans[102];
int total = 0;
vector<int>v[102];
int cmp(int p1,int p2) {
	if (w[p1] > w[p2])
		return 1;
	return 0;
}
void dfs(int root, int cnt) {
	total += w[root];
	ans[cnt] = w[root];
	if (total == s && v[root].size()==0) { //路径长度为s 并且root为叶子节点
		for (int i = 0; i <= cnt; i++) {
			if (i == 0)
				cout << ans[i];
			else
			cout << " " << ans[i];
		}
		cout << endl;
		return;
	}
	for (int i = 0; i < v[root].size(); i++) {
		dfs(v[root][i], cnt + 1);
		total -= w[v[root][i]];
	}
}
int main() {
	cin >> n >> m >> s;
	for (int i = 0; i < n; i++) {
		cin >> w[i];
	}
	for (int i = 0; i < m; i++) {
		int id, t, p;
		cin >> id >> t;
		for (int j = 0; j < t; j++) {
			cin >> p;
			v[id].push_back(p);
		}
		sort(v[id].begin(), v[id].end(), cmp);
	}
	dfs(0,0);
	return 0;
}