【PAT】A1053 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:

总结：

用静态树的构造逻辑上还是比较清晰的，就是一个数组，然后数组下有若干个子结点，通过子结点的数值找到数组中下标
这里的树和二叉树的不同之处在于，二叉树只有两个子结点，而树的子结点是不确定的，基于此，在构造结构的时候把子结点构造为用vector容器来存储，并用push_back和pop_back来插入删除
一般来说，输出的时候是要排序的，可能按结点的权重从大到小或者从小到大，本题给出一种思路，就是在输入的时候就把子结点按权重按要求排好序，这样在输出的时候就可以直接输出了。
在DFS中，首先要设置边界
一般来说，要有一个记录路径多少的数，这是为了输出方便，因为数与数之间需要用空格隔开，但是最后一个数后面不许出现空格。这就要求要做一个判别，这里用sumNode来判别的
DFS是一个递归的结构

代码：

#include <stdio.h>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn=110;
int total,row,sum;
int path[maxn];  //路径
struct Node{int data;vector<int> child;
}node[maxn];bool cmp(int a,int b){  //提前排序，有助于按降序输出 return node[a].data>node[b].data;
}
void DFS(int index,int sumNode,int s){  //结点，路径中的个数，和 if(s>sum) return;if(s==sum){if(node[index].child.size()!=0) return;for(int i=0;i<sumNode;i++){printf("%d",node[path[i]].data);if(i<sumNode-1) printf(" ");else printf("\n");}return;}for(int i=0;i<node[index].child.size();i++){int child=node[index].child[i];path[sumNode]=child;DFS(child,sumNode+1,s+node[child].data);}
}
int main(){scanf("%d%d%d",&total,&row,&sum);for(int i=0;i<total;i++){scanf("%d",&node[i].data);}int id,num,child;for(int i=0;i<row;i++){scanf("%d %d",&id,&num);for(int j=0;j<num;j++){scanf("%d",&child);node[id].child.push_back(child);}sort(node[id].child.begin(),node[id].child.end(),cmp);  //先对整个子树按权重从左到右由大向小排序，那之后的输出就不必再作比较了 }path[0]=0;//存放结点DFS(0,1,node[0].data); return 0;
}