PAT 1004 Counting Leaves

1004 Counting Leaves

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

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

救命， 不会写代码呀！！！

还是看看大佬的代码吧

题意：这题就是要你求出每一层中叶节点的个数

数据结构：

        （1）利用 vector二维数组 来存储每个节点的子节点，有点类似于邻接表，好处是可以直接使用 stl 自带的函数 .size() 直接计算出数组的大小，从而判断该节点是否为叶节点

        （2）数组 book 存放每一层叶节点的个数

思路：dfs 算法遍历每一个，可以在函数参数中假如变量表示当前所在的层数，假如当前点是叶节点，就在该层加一表示在该层发现了一个新的叶节点

dfs:

#include 
#include 
#include 
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);
    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);//因为 
    printf("%d", book[0]);
    for(int i = 1; i <= maxdepth; i++)
        printf(" %d", book[i]);
    return 0;
}

bfs：在求第几层的地方实质上就是求当前点到根节点的路劲（假设每个子节点和父节点的距离都为1）

#include 
#include 
#include 
using namespace std;
 
const int N=110;
int book[N],level[N];
int n,m,M=-1;
vector<int> v[N];
 
void bfs(){
    queue<int> q;
    q.push(1);
    level[1]=0;
    
    while(q.size()){
        int t=q.front();
        q.pop();
        
        if(v[t].size()==0){
            M=max(M,level[t]);
            book[level[t]]++;
        }
        for(int i=0;isize();i++){
            q.push(v[t][i]);
            level[v[t][i]]=level[t]+1;
        }
    }
}
 
int main(){
    cin >> n >> m;
    
    int k,node,r;
    for(int i=0;i
        scanf("%d%d",&node,&k);
        for(int j=0;j
            scanf("%d",&r);
            v[node].push_back(r);
        }
    }
    
    bfs();
    printf("%d",book[0]);
    for(int i=1;i<=M;i++)   printf(" %d",book[i]);
    
    return 0;
}

好好学习，天天向上！

我要考研！