http://www.patest.cn/contests/pat-a-practise/1094
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
这道题是2015考研机试前的那个PAT的D题 http://www.patest.cn/contests/pat-a-101-125-1-2015-03-14
那次PAT据说比机试简单。。。于是很多高分大神申请了免机试。。。性价比超高。。。。
我表示很忧伤。。。虽然我对自己机试成绩还算满意，但是还是有点感慨。。。如果参加这次PAT 说不定分数更好。。。嗯，做梦的感觉好好好哦。。。
分析：这道题需要根据指定序号成员的孩子信息（节点和节点的孩子信息）构造宗谱(树)，然后找到人数（节点）最多的那一代（层）。
1. N (<100) which is the total number of family members in the tree 所以最简单暴力的矩阵表示妥妥的时空都够用。
2. find the generation with the largest population 人数最多的那一代，即节点最多的那一层，层序遍历即可
总结：本题目的考点就是简单的层序遍历
1 #include<cstdio> 2 3 int main() 4 { 5 int family[100][100]={0}; //从1开始 0： 存放孩子数 For the sake of simplicity, let us fix the root ID to be 01 6 int generation[150]={0}; // 当前处理的代数的人员队列 从1开始 0:此代几个人 7 8 int members=0,havechildren=0,father=0,numchild=0,child=0; 9 scanf("%d%d",&members,&havechildren);// the total number of family members in the tree ----------------the number of family members who have children 10 for(int i=0;i<havechildren;i++) // ID K ID[1] ID[2] ... ID[K] 11 { 12 scanf("%d %d",&father,&numchild); 13 family[father][0]=numchild; 14 for(int j=1;j<=numchild;j++) //K (>0) is the number of his/her children 15 { 16 scanf("%d",&child); // ID's of his/her children 17 family[father][j]=child; 18 } 19 } 20 21 generation[1]=1;//the root level is defined to be 1 22 generation[0]=1; 23 int igenerat=1,largest=1,largestigenerat=1,start=2,len=1; 24 while(generation[0])//find the generation with the largest population 25 { 26 len=generation[0]; 27 start=len+1; 28 generation[0]=0; 29 for(int i=1;i<=len;i++) 30 { 31 father=generation[i]; //当前处理的father 32 numchild=family[father][0]; 33 for(int j=1;j<=numchild;j++) 34 { 35 generation[start]=family[father][j]; 36 generation[0]++; 37 start++; 38 } 39 }//zhengli shuzu 40 41 for(int i=1;i<=generation[0];i++) generation[i]=generation[i+len]; 42 igenerat++; 43 if(generation[0]>largest) largest=generation[0],largestigenerat=igenerat; 44 } 45 46 printf("%d %d",largest,largestigenerat); //the largest population number and the level of the corresponding generation 47 return 0; 48 49 }