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<2
30 , 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 {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:
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
#include<cstdio>
#define maxn 102
#include<algorithm>
#include<vector>
#include<cmath>
#include<iostream>
#include<queue>
using namespace std;
//输入数据生成其对应的树，对树进行先根遍历(DFS)，每次将当前访问的结点的权值累加，并将结点加入答案数组
//如果当前权和等于给出的权值，进行"剪枝"，即到达递归边界，输出答案序列，如果大于则也到达递归边界
//树的子结点用vector存储，使用静态方法存储树，用下标代替指针，由于题目要求降序输出答案序列，故在子结点vecotr
//中，先排序，这样DFS过程总是先求出权和最大的答案序列
struct node
{
int w;
vector<int>child;
}Node[maxn];
vector<int>ans;
int S;
int sumw=0;
bool cmp(int a,int b)
{
return Node[a].w>Node[b].w;
}
void preorder(int root)
{
sumw+=Node[root].w;
ans.push_back(Node[root].w);
if(sumw==S){
if(Node[root].child.size()==0){
//是根节点
for(int i=0;i<ans.size();i++){
if(i!=ans.size()-1){
printf("%d ",ans[i]);
}
else{
printf("%d\n",ans[i]);
}
}
sumw-=Node[root].w;
ans.pop_back();
return ;
}
else{
sumw-=Node[root].w;
ans.pop_back();
return ;
}
}
if(sumw>S){
sumw-=Node[root].w;
ans.pop_back();
return ;
}
for(int i=0;i<Node[root].child.size();i++){
preorder(Node[root].child[i]);
}
sumw-=Node[root].w;
ans.pop_back();
}
int main()
{
int N,M;
scanf("%d %d %d",&N,&M,&S);
for(int i=0;i<N;i++){
scanf("%d",&Node[i].w);
}
while(M--){
int index,i;
scanf("%d %d",&index,&i);
for(int j=0;j<i;j++){
int temp;
scanf("%d",&temp);
Node[index].child.push_back(temp);
}
sort(Node[index].child.begin(),Node[index].child.end(),cmp);
}
preorder(0);
return 0;
}