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, the number of nodes in a tree, M (<), the number of non-leaf nodes, and 0, the given weight number. The next line contains N positive numbers where Wi (<) 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 { is said to be greater than sequence { if there exists 1 such that Ai=Bi for ,, 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
静态建立树，并求根节点到叶节点的权重之和=s的所有情况,非递减输出所有权重情况；
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
#define maxn 105
struct Node
{
int data;
vector<int>child;
}node[maxn];
bool cmp(int a, int b)//对node里child数组里的元素进行排序，排序规则是node[a].data > node[b].data，这样可以做到非递增输出
{
return node[a].data > node[b].data;
}
int n, m, s;
int path[maxn];//自己写的用path直接保存了路径上的权重值，也可以直接保存路径的编号，大同小异
void BFS(int idex, int num, int sum)
{
if (sum > s)return;
if (sum == s )
{
if(node[idex].child.size() != 0)
return;
for (int i = 0; i < num; i++)
{
cout << path[i];
if (i != num-1)
cout << " ";
else
cout << endl;
}
return;
}
int len = node[idex].child.size();
for (int i = 0; i < len; i++)
{
int child = node[idex].child[i];
path[num] = node[child].data;
BFS(child, num+1,sum + node[child].data);
}
}
int main()
{
FILE* T;
freopen_s(&T, "input.txt", "r", stdin);
cin >> n >> m >> s;
int pos,cnt, child;
for (int i = 0; i < n; i++)
cin >> node[i].data;
for (int i = 0; i < m; i++)
{
cin >> pos >> cnt;
for (int j = 0; j < cnt; j++)
{
cin >> child;
node[pos].child.push_back(child);
}
sort(node[pos].child.begin(), node[pos].child.end(),cmp);
}
path[0] = node[0].data;
BFS(0, 1, node[0].data);
return 0;
} 
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