1004 Counting Leaves （30 分)

2019-03-05

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:

1	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:

1 2	2 1 01 1 02

Sample Output:

0 1

作者: CHEN, Yue

单位: 浙江大学

时间限制: 400 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

给出n,m代表树的节点数以及非叶结点树，接下来输入m行，每行格式id k id[1] id[2]..代表id结点的k个儿子。要求层序输出各层无儿子结点的数量。

分析

利用vector<int>数组存储各结点的儿子号，利用lay数组记录结点所在的层数，利用bfs输出结果

代码

#include <iostream>
#include <vector>
#include <algorithm>
#include <deque>
#include <map>
#include <cstring>
using namespace std;

int main() {
    int n,m,cur,k,t,maxlay=-1;
    int lay[110]={},res[110]={};
    cin>>n>>m;
    if(n==0)
        return 0;
    vector<int> vec[110];
    for(int i=0;i<m;i++){
        cin>>cur>>k;
        for(int j=0;j<k;j++){
            cin>>t;
            vec[cur].push_back(t);
        }
    }
    lay[1]=0;
    deque<int> que;
    que.push_back(1);
    while(!que.empty()){
        cur=que.front();
        que.pop_front();
        //记录树的深度
        maxlay=max(maxlay,lay[cur]);
        if(vec[cur].empty()){//该结点无儿子
            res[lay[cur]]++;
        }
        else
        for(int a:vec[cur]){//将该结点的儿子入队，儿子的层数=父亲层数+1
            que.push_back(a);
            lay[a]=lay[cur]+1;
        }
    }
    cout<<res[0];
        for(int i=1;i<=maxlay;i++)
            cout<<" "<<res[i];
    return 0;
}