PAT 1094.The Largest Generation

1094. The Largest Generation (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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
题目分析：
01一定是家族谱系的根层级为1，以此开始做bfs搜索，更新每个节点的层级即可.用了队列完成bfs。本题无难度

#include<stdio.h>
#include<vector>
#include<queue>

using namespace std;

class Member{
public:
    int level; 
    vector<int> children;
};

Member mem[100];
queue<int> now;
int level = 1;
int sum[100];

void bfs(int start){

}

int main(void){
    int n, m;
    scanf("%d %d", &n, &m);
    int i, j;
    int parent;
    int child, cnum;
    for (i = 0; i < m; i++){
        scanf("%d %d", &parent, &cnum);
        for (j = 0; j < cnum; j++){
            scanf("%d", &child);
            mem[parent].children.push_back(child);
        }
    }
    mem[1].level = 1;
    sum[1] = 1;
    now.push(1);
    while (now.size() != 0){
        parent = now.front();
        for (i = 0; i < mem[parent].children.size(); i++){
            child = mem[parent].children[i];
            mem[child].level = mem[parent].level + 1;
            sum[mem[child].level]++;
            now.push(child);
        }
        now.pop();
    }
    int max = 1;
    for (i = 1; i <= n; i++){
        if (sum[max] < sum[i]){
            max = i;
        }
    }
    printf("%d %d\n", sum[max], max);
    return 0;
}