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 18Sample 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
1 #include<cstdio> 2 #include<algorithm> 3 #include<iostream> 4 #include<cstring> 5 #include<queue> 6 #include<vector> 7 #include<cmath> 8 using namespace std; 9 vector<int> v[100]; 10 int main(){ 11 int n,m; 12 scanf("%d %d",&n,&m); 13 int i,num,count,in; 14 for(i=0;i<m;i++){ 15 scanf("%d %d",&num,&count); 16 while(count--){ 17 scanf("%d",&in); 18 v[num].push_back(in); 19 } 20 } 21 queue<int> q; 22 q.push(1); 23 int last=1,tail,floor=1,cur,maxcount=1,maxfloor=1,curcount=0; 24 while(!q.empty()){ 25 cur=q.front(); 26 q.pop(); 27 for(i=0;i<v[cur].size();i++){ 28 q.push(v[cur][i]); 29 tail=v[cur][i]; 30 curcount++; 31 } 32 if(last==cur){ 33 last=tail; 34 floor++;//point to next floor 35 if(curcount>maxcount){ 36 maxcount=curcount; 37 maxfloor=floor; 38 } 39 curcount=0; 40 } 41 } 42 printf("%d %d ",maxcount,maxfloor); 43 return 0; 44 }