1094. The Largest Generation (25)
A family hierarchy is usually presented by a pedigree tree where all the nodes on the same level belong to the same generation. Your task is to find the generation with the largest population.
Input Specification:
Each input file contains one test case. Each case starts with two positive integers N (<100) which is the total number of family members in the tree (and hence assume that all the members are numbered from 01 to N), and M (<N) which is the number of family members who have children. Then M lines follow, each contains the information of a family member in the following format:
ID K ID[1] ID[2] ... ID[K]
where ID is a two-digit number representing a family member, K (>0) is the number of his/her children, followed by a sequence of two-digit ID's of his/her children. For the sake of simplicity, let us fix the root ID to be 01. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print in one line the largest population number and the level of the corresponding generation. It is assumed that such a generation is unique, and the root level is defined to be 1.
Sample Input:
23 13
21 1 23
01 4 03 02 04 05
03 3 06 07 08
06 2 12 13
13 1 21
08 2 15 16
02 2 09 10
11 2 19 20
17 1 22
05 1 11
07 1 14
09 1 17
10 1 18
Sample Output:
9 4
姥姥教的方法，顺便用一下。
last：更新前，本层最后一个元素
tail：下一层当前最后一个元素
floor：指向当前的层数
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<queue>
#include<vector>
#include<cmath>
using namespace std;
vector<int> v[];
int main(){
int n,m;
scanf("%d %d",&n,&m);
int i,num,count,in;
for(i=;i<m;i++){
scanf("%d %d",&num,&count);
while(count--){
scanf("%d",&in);
v[num].push_back(in);
}
}
queue<int> q;
q.push();
int last=,tail,floor=,cur,maxcount=,maxfloor=,curcount=;
while(!q.empty()){
cur=q.front();
q.pop();
for(i=;i<v[cur].size();i++){
q.push(v[cur][i]);
tail=v[cur][i];
curcount++;
}
if(last==cur){
last=tail;
floor++;//point to next floor
if(curcount>maxcount){
maxcount=curcount;
maxfloor=floor;
}
curcount=;
}
}
printf("%d %d\n",maxcount,maxfloor);
return ;
}