PAT

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 2410 2 4 3 5 10 2 18 9 7 2 2 1 3 12 1 8 6 2 200 4 01 02 03 0402 1 0504 2 06 0703 3 11 12 1306 1 0907 2 08 1016 1 1513 3 14 16 1717 2 18 19

Sample Output:

10 5 2 710 4 1010 3 3 6 210 3 3 6 2

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给定条件：
1.n个节点和节点的权重
2.节点间连接

要求：
1.从根节点出发到叶子节点的路线中节点权重之和是给定值的路线
2.存在多条路线，根据给定的要求按照顺序输出

解决：
1.从根结点开始搜索，搜索边界为（1找到叶子节点 2当前和已超过给定数值）
2.如果搜索到叶子节点要判断当前和是否等于给定值。如果符合则该路线是符合条件的
3.将搜索到的路线存放到容器中
4.将路线按照给定要求进行排序，然后输出结果

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#include<cstdio>#include<vector>#include<algorithm>using namespace std;int n, m, s, w[110];int father, k, son;bool vis[110];vector<int> e[110];vector<int> path;vector< vector<int> > ans;void dfs(int cur, int sum){if(e[cur].size() == 0){if(sum == s){ans.push_back(path);}return ;}if(sum > s){return ;}for(int i = 0; i < e[cur].size(); i++){int next = e[cur][i];if(vis[next] == false){vis[next] = true;path.push_back(w[next]);dfs(next, sum+w[next]);path.pop_back();vis[next] = false;}}}int main(){while(scanf("%d%d%d",&n, &m, &s) != EOF){fill(vis, vis+110, false);// inputfor (int i = 0; i < n; i++){scanf("%d",&w[i]);}for (int i = 0; i < m; i++){scanf("%d%d",&father, &k);for(int i = 0; i < k; i++){scanf("%d", &son);e[father].push_back(son);}}// dfsvis[0] = true;path.push_back(w[0]);dfs(0,w[0]);// outputsort(ans.begin(),ans.end());for(int i = ans.size()-1; i >= 0; i--){for(int j = 0; j < ans[i].size(); j++){if(j!=0) printf(" ");printf("%d", ans[i][j]);}printf("\n");}}return 0;}