1053 Path of Equal Weight

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 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<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
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

dfs，答案要先保存，经过排序后再输出。 

#include <iostream>
#include <iomanip>
#include <vector>
#include <algorithm>
using namespace std;
int a[1000];//存储每个节点的权值
bool vis[1000];
int ans[1000];
int n, m, num, x, y, k;

struct Array {
	int size;
	int a[1000];
} b[1000];
int cnt;

vector<int>g[1000];//存储非叶节点的孩子
bool cmp(Array a1, Array a2) {
	int t = min(a1.size, a2.size);
	for (int i = 1; i <= t; i++) {
		if (a1.a[i] > a2.a[i]) {
			return true;
		} else if (a1.a[i] < a2.a[i]) {
			return false;
		}
	}
	return false;//这句必须有，不然会出现段错误
}

void dfs(int u, int sum, int ind) {
	if (sum + a[u] == num && !(g[u].size() > 0)) {
		ans[ind] = a[u];
		for (int i = 1; i <= ind; i++) {
			b[cnt].a[i] = ans[i];
			b[cnt].size = ind;
		}
		cnt++;
		return;
	}
	if (sum + a[u] < num) {
		sum += a[u];
		ans[ind] = a[u];
		if (g[u].size() > 0) {
			for (int j = 0; j < g[u].size(); j++) {
				if (!vis[g[u][j]]) {
					vis[g[u][j]] = 1;
					dfs(g[u][j], sum, ind + 1);
					vis[g[u][j]] = 0;
				}
			}
		}
	}
}

int main() {
	cin >> n >> m >> num;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
	}
	for (int i = 0; i < m; i++) {
		cin >> x >> k;
		for (int j = 0; j < k; j++) {
			cin >> y;
			g[x].push_back(y);
		}
	}
	vis[0] = 1;
	dfs(0, 0, 1);
	sort(b, b + cnt, cmp);
	for (int i = 0; i < cnt; i++) {
		for (int j = 1; j <= b[i].size; j++) {
			cout << b[i].a[j];
			if (j != b[i].size) {
				cout << ' ';
			}
		}
		if (i != cnt - 1) {
			cout << endl;
		}
	}
	return 0;
}