PAT_A 1053. Path of Equal Weight (30)

1053. Path of Equal Weight (30)

Given a non-empty tree with root R, and with weight Wi assigned to each tree node Ti. Theweight of a path from R to L is defined to be the sum of the weights of all the nodes alongthe 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 theweight 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 62}, which correspond to the red edges in Figure 1.Figure 1Input 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 itschildren, followed by a sequence of two-digit ID's of its children. For the sake ofsimplicity, 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 pathoccupies 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 2410 2 4 3 5 10 2 18 9 7 2 2 1 3 12 1 8 6 2 200 4 01 02 03 0402 1 0504 2 06 0703 3 11 12 1306 1 0907 2 08 1016 1 1513 3 14 16 1717 2 18 19Sample Output:10 5 2 710 4 1010 3 3 6 210 3 3 6 2

• 分析：

  • 题目：
    
    • 给定一棵树，每个节点有一个权值，给定一个权值和S,找到所有从根到叶子的权值和为S的路径。
    • 输出要求，不同路径之间要求按非降顺序输出。
    • 考树的构造，遍历。
  • 解题：
    
    • 先构造这棵树，这里用vector tree[110]表示，vector中记录孩子节点号。每个节点编号对应的权值，
      　 nodeWeight[110]。
    • 由于结果要求非降序输出，故对输入的数据进行预处理使其有序。对于每个节点的还在按权值递增排序存储在vector元素中，这样遍历树时有序遍历（从大到小)，到达叶子，满足输出，便是符合排序要求。

• code:

#include<iostream>#include<cstdio>#include<algorithm>#include<vector>using namespace std;int nodeWeight[110];vector<int> tree[110];int check[110];bool comp(int a,int b){    return nodeWeight[a]<nodeWeight[b];}//右递归下去vector<int> seq;void checkTree(int child,long w,long s,vector<int>& seq){    if(child>=110)return;    if(w>s)return;    if(tree[child].size()==0)    {        //leaf        if(w==s)        {          for(int k=0;k<seq.size()-1;k++)          {              printf("%d ",seq[k]);          }          printf("%d\n",seq[seq.size()-1]);          return;        }        return;    }    for(int i=tree[child].size()-1;i>=0;i--)    {        w+=nodeWeight[tree[child][i]];        seq.push_back(nodeWeight[tree[child][i]]);        checkTree(tree[child][i],w,s,seq);        w-=nodeWeight[tree[child][i]];        seq.pop_back();    }}int main(){    freopen("in","r",stdin);    fill_n(nodeWeight,110,0);    fill_n(check,110,0);    int N,M,tmp;    long S;    scanf("%d%d%ld",&N,&M,&S);    for(int i=0;i<N;i++)    {        scanf("%d",&tmp);        nodeWeight[i]=tmp;    }    int count=0;    int child=0;    for(int i=0;i<M;i++)    {        scanf("%d%d",&tmp,&count);        for(int j=0;j<count;j++)        {            scanf("%d",&child);            tree[tmp].push_back(child);        }        //按w递增        sort(tree[tmp].begin(),tree[tmp].end(),comp);    }    seq.push_back(nodeWeight[0]);    checkTree(0,nodeWeight[0],S,seq);    return 0;}

• AC: