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

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 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

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 101using 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;}