1053 Path of Equal Weight

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. 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:

1
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:
1
2
3
4
5
6
7
8
9
10
11
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:
1
2
3
4
10 5 2 7
10 4 10
10 3 3 6 2
10 3 3 6 2

浅析
Q：找出从根节点到叶子结点权值和等于给定值的所有路径，并从大到小输出。

Code
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#include
#include
#include
using namespace std;

const int maxn = 110;
struct node{
int weight;
vector<int> child;
} TNode[maxn];
bool cmp(int a,int b){
return TNode[a].weight > TNode[b].weight;
}
int n, m, weight; //起始输入的结点树，非叶子节点数，目标值
int path[maxn];
/*尝试用vector来构建path，但是个人能力局限，不能很好的进行
push_back操作，有想法的可以自己改着试试*/
void DFS(int index,int num_node,int sum){
if(sum>weight)
return;
if(sum == weight){
if(TNode[index].child.size()!=0)
return;
for (int i = 0; i < num_node; i++)
{
printf("%d",TNode[path[i]].weight);
if(i-1)
printf(" ");
else
printf("\n");
}
return;
}
for (int i = 0; i < TNode[index].child.size();i++){
int child = TNode[index].child[i];
path[num_node] = child;
DFS(child, num_node+1,sum + TNode[child].weight);
}
}
int main(){
scanf("%d %d %d", &n, &m, &weight);
for (int i = 0; i < n;i++)
scanf("%d", &TNode[i].weight);
int father,k, child;
for (int i = 0; i < m;i++){
scanf("%d %d",&father,&k);
for (int j = 0; j < k;j++){
scanf("%d", &child);
TNode[father].child.push_back(child);
}
sort(TNode[father].child.begin(), TNode[father].child.end(), cmp);
}
path[0] = 0;
DFS(0, 1, TNode[0].weight);
return 0;
}