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:
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 |
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:
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1 |
9 4 |
与 1004 思路类似
代码如下：
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 |
//与 1004 思路类似 #include<stdio.h> int intree[100],tree[100],population[100]; int main() { int i,j,n,m,k,child,parent,level,temp,edge=0,max=0; scanf("%d%d",&n,&m); intree[1]=1; for(i=0;i<m;i++) { scanf("%d%d",&parent,&k); intree[parent]=1; for(j=0;j<k;j++) { scanf("%d",&child); intree[child]=1; tree[child]=parent; } } for(i=0;i<100;i++) { if(intree[i]==1) { level=0; temp=i; while(temp!=1) { level++; temp=tree[temp]; } if(level>edge) edge=level; population[level]++; } } for(i=0;i<=edge;i++) { if(population[i]>population[max]) { max=i; } } printf("%d %d",population[max],max+1); } |