1004 Counting Leaves

1004 Counting Leaves (30分)

A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.

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, and M (<N), the number of non-leaf nodes. 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 01.

The input ends with N being 0. That case must NOT be processed.

Output Specification:

For each test case, you are supposed to count those family members who have no child for every seniority level starting from the root. The numbers must be printed in a line, separated by a space, and there must be no extra space at the end of each line.

The sample case represents a tree with only 2 nodes, where 01 is the root and 02 is its only child. Hence on the root 01 level, there is 0 leaf node; and on the next level, there is 1 leaf node. Then we should output 0 1 in a line.

Sample Input:

2 1
01 1 02
​
      
        
      
    

Sample Output:

0 1

审题

题目给出一棵树，我们需要求出该树每层的叶子结点。

思路

这里用BFS的方法把每层的叶结点求出存值数组，然后按层序输出。

参考代码

#include<iostream>
#include<stdio.h>
#include<vector>
#include<queue>
using namespace std;
​
typedef struct {
    vector<int> child;
}Node;
​
Node node[100];
int level[100];
int n,m;
​
void BFS(int root)
{
    queue<int> que;
    que.push(root);
    int last=1;
    int count=0;
    int l=0;
    while(!que.empty())
    {
        int front=que.front();
        que.pop();
        if(node[front].child.size()==0){
            count++;
        } else {
            for(int i=0;i<node[front].child.size();i++){
                que.push(node[front].child[i]);
            }
        }
        //第一次提交的代码，逻辑混乱！！
        //int num=node[front].child.size();
        //  if(num==0){
        //    level[l]=count;
        //      count=0;
        //      l++;
        //  } else { 
        //      level[l]=count;
        //      last=que.back();
        //      count=0;
        //      l++;
        //  }
        if(front==last){
                level[l]=count;
                last=que.back();
                count=0;
                l++; 
        }
    }
    cout<<level[0];
    for(int i=1;i<l;i++)
    {
        cout<<" "<<level[i];
    }
}
​
int main()
{
    while(cin>>n)
    {
        if(n!=0){
            cin>>m;
            for(int i=0;i<m;i++)
                {
                    int id;
                    int num;
                    cin>>id>>num;
                    for(int j=0;j<num;j++)
                    {
                        int c;
                        cin>>c;
                        node[id].child.push_back(c); 
                    }
                }
                BFS(1);
        } else {
            break;
        }
    }
    
}

 

总结

本题是层序遍历的模板题。第一次的代码逻辑混乱，花了好长时间才检查出来，看来写代码时的专注度不高，导致了逻辑漏洞，同时我发现逻辑越混乱检查难度也越大。