7-4树叶

A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.
家庭等级通常由谱系树表示。您的工作是计算没有孩子的家庭成员。

Input Specification:

The input consists of several test cases, each 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.
输入规格：
输入由几个测试用例组成，每个用例都从包含0 <N <100（一行在树中的节点数）和M（<N）（非叶节点的数）的行开始。然后，接着M行，每行的格式为ID K ID [1] ID [2] … ID [K]，其中ID是代表给定非叶节点的两位数字，K是其子代的数目，随后是其子代的两位数字ID序列。为了简单起见，让我们将根ID固定为01。

输入以N为0结尾。不得处理这种情况。

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.

For example, the first 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.
输出规格：
对于每个测试用例，应该从根开始计算每个资历级别中没有孩子的家庭成员。数字必须成行打印，并用空格隔开，并且每行末尾不得有多余的空格。

例如，第一个样本案例表示只有2个节点的树，其中01是根，而02是其唯一的子节点。因此，在根01级别上，存在0个叶节点；在下一个级别上，有1个叶节点。然后，我们应该在一行中输出0 1。

Sample Input:

2 1
01 1 02
1 0
7 4
01 2 02 03
06 1 07
02 2 04 05
03 1 06
0 0

Sample Output:

0 1
1
0 0 2 1

解析

这题的意思是每一层有几个叶节点，当输入7 4 时

DFS

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

vector<int> v[100];
int book[100], maxdepth = -1;

void dfs(int index, int depth) {
    if(v[index].size() == 0) {
        book[depth]++;
        maxdepth = max(maxdepth, depth);
        return ;
    }
    for(int i = 0; i < v[index].size(); i++){
        dfs(v[index][i], depth + 1);
        
    }
        
}
int main() {
    int n, m, k, node, c;
    scanf("%d %d", &n, &m);
    while(true){
        if(n==0&&m==0)
            return 0;
        for(int i=0;i<100;i++){
            book[i]=0;
            v[i].clear();
        }
            
        maxdepth=-1;
        
        for(int i = 0; i < m; i++) {
            scanf("%d %d",&node, &k);
            for(int j = 0; j < k; j++) {
                scanf("%d", &c);
                v[node].push_back(c);
            }    
        }
        dfs(1, 0);
        printf("%d", book[0]);
        for(int i = 1; i <= maxdepth; i++)
            printf(" %d", book[i]);
        cout<<endl;
        scanf("%d %d", &n, &m);
    }
    
    return 0;
}

数组实现

#include<bits/stdc++.h>
using namespace std;
int a[105][105];
int book[105];
int maxdepth=-1;

void def(int index,int deptf){
    if(a[index][0]==0){
        book[deptf]++;
        if(deptf>=maxdepth){
            maxdepth=deptf;
        }
        return ;
    }
    for(int i=1;i<=a[index][0];i++){
        def(a[index][i],deptf+1);
    }
}
int main(){
    int n,m;
    
    cin>>n>>m;
    while(true){
        for(int i=0;i<105;i++){
            book[i]=0;
            for(int j=0;j<105;j++){
                a[i][j]=0;
            }
        }
        maxdepth=-1;
        if(n==0&&m==0)
            break;
        int x,y,z;
        for(int i=1;i<=m;i++){
            cin>>x>>y;
            a[x][0]=y;
            for(int j=1;j<=y;j++){
                cin>>z;
                a[x][j]=z;
            }        
        }
        def(1,0);
        cout<<book[0];
        for(int i=1;i<=maxdepth;i++){
            cout<<" "<<book[i];
        }
        cout<<endl;
        cin>>n>>m;
    }
    return 0;
}