PAT甲级1053、 Path of Equal Weight

题目

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 $A_i$ = $B_i$ 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:

思路

本来没想写的，但是写完这个题快吃饭了，不知道干点什么，就来水一篇吧

思路很简单

先给权重后给子节点，用链表的话需要二次操作才能建树，而且涉及到了ID，放在结构体里面肯定不合适，所以考虑到结构体数组或者说静态树。

另外说一下，总结点不过100，量级比较小，冗余的话也不要紧

然后另一个要点的话就是DFS或者说回溯算法，都差不多，去遍历这棵树寻找路径

这里的栈有点多余了，用一个vector代替的话应该会更好一点

#include <iostream>
#include <stack>
#include <vector>
#include <algorithm>
using namespace std;
struct node
{int weight;vector<int> children;
}nodes[105];
vector<vector<int>> paths;
bool cmp(vector<int> a, vector<int> b)
{int l = min(a.size(),b.size());int i;for(i=0;i<l;i++){if(a[i] != b[i])break;}return a[i] > b[i];
};
stack<int> path;
vector<int> pathWeight;
int s,weightSum = 0;
void DFS(int cur)
{path.push(cur);pathWeight.push_back(nodes[cur].weight);weightSum += nodes[cur].weight;if(nodes[cur].children.empty() && weightSum == s) paths.push_back(pathWeight);for(int i=0;i<nodes[cur].children.size();i++){DFS(nodes[cur].children[i]);}path.pop();pathWeight.pop_back();weightSum -= nodes[cur].weight;return ;
}int main()
{int n,m;cin>>n>>m>>s;for(int i=0;i<n;i++){cin>>nodes[i].weight;}for(int i=0;i<m;i++){int id,k;cin>>id>>k;for(int j=0;j<k;j++){int Id;cin>>Id;nodes[id].children.push_back(Id);}}// for(int i=0;i<n;i++)// {//     cout<<i<<" "<<nodes[i].weight<<" "<<nodes[i].children.size();//     for(int j=0;j<nodes[i].children.size();j++)//         cout<<" "<<nodes[i].children[j];//     cout<<endl;// }DFS(0);sort(paths.begin(), paths.end(),cmp);for(int i=0;i<paths.size();i++){cout<<paths[i][0];for(int j=1;j<paths[i].size();j++)cout<<" "<<paths[i][j];cout<<endl;}
}