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 , the number of nodes in a tree, and (), the number of non-leaf nodes. Then 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 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
代码实现:
#include<iostream>
#include<queue>
#include<vector>
using namespace std;
const int N=1e5+5;
vector<int>v[N];
int level[N];
int idx;
void bfs(int root){
queue<pair<int,int> >q;
q.push({root,1});
while(q.size()){
int s=q.front().first;
int dis=q.front().second;
idx=max(idx,dis);
q.pop();
if(v[s].size()==0){
level[dis]++;
continue;
}
for(int i=0;i<v[s].size();i++){
q.push({v[s][i],dis+1});
}
}
}
int main(){
int n,m;
cin>>n>>m;
while(m--){
string s;
int k;
cin>>s>>k;
while(k--){
string s1;
cin>>s1;
int x=stoi(s);
int y=stoi(s1);
v[x].push_back(y);
}
}
bfs(1);
for(int i=1;i<=idx;i++){
if(i==1)cout<<level[i];
else cout<<" "<<level[i];
}
return 0;
}