1094. The Largest Generation (25)

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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 1321 1 2301 4 03 02 04 0503 3 06 07 0806 2 12 1313 1 2108 2 15 1602 2 09 1011 2 19 2017 1 2205 1 1107 1 1409 1 1710 1 18

Sample Output:

9 4

-------------------华丽的分割线-------------------

分析：还是可以用一个二维数组来存输入信息。Graph[i][j] != 0表示j是i的子节点。

然后进行广度优先搜索就好了。

代码：

#include<cstdio>#include<cstdlib>#include<queue>using namespace std;#define Maxn 101 int Graph[Maxn][Maxn]; int level[Maxn]; int result[Maxn]; int N; int M;void BFS(int start);int main(void){scanf("%d %d\n",&N,&M);int i,j,k;int parent,child;for(i=0;i<M;++i){scanf("%2d %d",&parent,&k);for(j=0;j<k;++j){scanf("%2d",&child);Graph[parent][child] = 1;}}BFS(1);for(i=1;i<=N;++i){++result[level[i]];}int max = 0;int maxlevel = 0;for(i=1;i<=N;++i){if(result[i] > max){max = result[i];maxlevel = i;}}printf("%d %d",max,maxlevel);system("pause");return 0;}void BFS(int start){queue<int> myque;myque.push(start);level[start] = 1;int thisone;int i;while(!myque.empty()){thisone = myque.front();myque.pop();for(i=1;i<=N;++i){if(Graph[thisone][i] != 0){level[i] = level[thisone] + 1;myque.push(i);}}}}

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