PAT(A) - 1094. The Largest Generation (25)

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本文介绍了一种通过构建家谱树并使用广度优先搜索(BFS)来确定家族成员中人口最多的一代的方法。输入包含家庭成员数量及子女信息，输出最大人口数及其对应层级。

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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

思路分析：建立一棵树（普通的树）表示家谱，进行一次BFS，并在遍历的时候统计每层结点的数目，最后输出最多层的结点数目和层号。

#include <cstdio>
#include <vector>
#include <queue>

#define MAX 101

using namespace std;

typedef struct {
    int index;
    vector<int> child;
    int level;
} Node;

Node node[MAX];
int countLevel[MAX] = { 0 };

int main() {
    //freopen( "123.txt", "r", stdin );
    int n, m;
    scanf( "%d%d", &n, &m );

    int id, k;
    for( int i = 1; i <= m; i++ ) {
        scanf( "%d%d", &id, &k );
        node[id].index = id;
        int childID;
        for( int j = 0; j < k; j++ ) {
            scanf( "%d", &childID );
            node[id].child.push_back( childID );
        }
    }

/*
    for( int i = 1; i <= n; i++ ) {
        printf( "%d node: ", i );
        int num = node[i].child.size();
        for( int j = 0; j < num; j++ ) {
            printf( "%d ", node[i].child[j] );
        }
        printf( "\n" );
    }
*/

    int root = 1;
    node[root].level = 1;

    int maxCount = 0;
    int maxLevel = 0;
    queue<Node> q;
    q.push( node[root] );
    while( !q.empty() ) {
        Node curNode = q.front();
        countLevel[curNode.level]++;
        if( countLevel[curNode.level] > maxCount ) {
            maxCount = countLevel[curNode.level];
            maxLevel = curNode.level;
        }
        q.pop();
        int k = curNode.child.size();
        for( int i = 0; i < k; i++ ) {
            node[curNode.child[i]].level = curNode.level + 1;
            q.push( node[curNode.child[i]] );
        }
    }

    printf( "%d %d", maxCount, maxLevel );
    return 0;
}