https://pintia.cn/problem-sets/994805342720868352/problems/994805521431773184
A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.
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.
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.
hierarchy 等级制度
pedigree tree 谱系树
non-leaf nodes 非叶节点
format 格式
the sake of simplicity 为了简单
for every seniority level 每个资历级别
#include<bits/stdc++.h> #pragma GCC optimize(3) #define max(a,b) a>b?a:b using namespace std; typedef long long ll; vector<int> v[105]; int deep[105]; int ans[105]; int main(){ int n,m; scanf("%d%d",&n,&m); for(int i=1;i<=m;i++){ int a,b,k; scanf("%d%d",&a,&k); for(int j=1;j<=k;j++){ scanf("%d",&b); v[a].push_back(b); } } deep[1]=1; queue<int> q; q.push(1); int up=1; while(!q.empty()){ int fa=q.front(); q.pop(); if(v[fa].empty()){ ans[deep[fa]]++; continue; } for(int i=0;i<v[fa].size();i++){ deep[v[fa][i]]=deep[fa]+1; up=max(up,deep[v[fa][i]]); q.push(v[fa][i]); } } for(int i=1;i<=up;i++) printf("%d%c",ans[i]," \n"[i==up]); return 0; }
