1053. Path of Equal Weight (30)

最新推荐文章于 2022-04-13 00:26:50 发布

原创最新推荐文章于 2022-04-13 00:26:50 发布 · 503 阅读

0 ·

CC 4.0 BY-SA版权

PAT 专栏收录该内容

79 篇文章

订阅专栏

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 Figure 1: 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 Figure 1.

Figure 1

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 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 {A₁, A₂, ..., A_n} is said to be greater than sequence {B₁, B₂, ..., 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:

思路：很具代表性的一道图的遍历的题目，用DFS和邻接链表的数据结构，基本没有什么坑点，但是题目要读仔细，特别是加粗的那几个字

AC参考代码：

#include <iostream>
#include <vector>
#include <fstream>
#include <algorithm>
using namespace std;

int N,M,S;
vector<int> node;
vector<vector<int> > edge(101);//edge中储存的是索引 索引再根据node找到对应的值
vector<vector<int> >result;//把符合条件的路径节点作为vector数组储存起来
vector<int> isVisited(101);//标记节点是否被遍历
vector<int> currentNode;//储存当前路径上的节点
int sum = 0;//路径权重和

int compare(vector<int> a,vector<int> b){//按照字典逆序排序
    int minN = a.size()<b.size()?a.size():b.size();
    for(int i=0;i<minN;i++){
        if(a[i]!=b[i]){
            return a[i]>b[i];
        }
    }
    return 0;
}
void DFS(int i){    //遍历并寻找满足条件的路径并记录节点
    isVisited[i] = 1;
    currentNode.push_back(node[i]);
    sum += node[i];
    if(edge[i].size()==0){//到达叶子节点
        if(sum==S){ //这条路径符合要求
            result.push_back(currentNode);
        }
    }else{
        for(int j=0;j<edge[i].size();j++){
            DFS(edge[i][j]);
        }
    }
    sum -= node[i];     //遍历完叶子节点后返回需要修改当前路径节点和节点和
    currentNode.pop_back();
}
int main()
{
    //ifstream cin("1.txt");
    cin>>N>>M>>S;
    node.resize(N);
    for(int i=0;i<N;i++){
        cin>>node[i];
    }
    for(int i=0;i<M;i++){
        int index,arrSize;
        cin>>index>>arrSize;
        for(int j=0;j<arrSize;j++){
            int temp;
            cin>>temp;
            edge[index].push_back(temp);
        }
    }
    DFS(0);
    sort(result.begin(),result.end(),compare);//按字典逆序排序
    for(int i=0;i<result.size();i++){
        for(int j=0;j<result[i].size();j++){
            j<result[i].size()-1?cout<<result[i][j]<<" ":cout<<result[i][j]<<endl;
        }
    }
    return 0;
}