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

   层次遍历的非递归算法，刚写一会儿就想到了，但是不熟悉，写了一个来小时。。

#include<iostream>#include<vector>using namespace std;typedef struct node{public:    vector <int>child;}* tree;int N,M;void order(vector <tree> tr);int main(){    freopen("in.txt","r",stdin);    cin>>N>>M;    vector <tree> family(N+1);    for(int i=1;i<=N;i++)        family[i]=new node();    for(int i=0;i<M;i++)    {        int f,k,c;        cin>>f>>k;        while(k--)        {            cin>>c;            family[f]->child.push_back(c);        }    }    order(family);    system("pause");    return 0;}void order(vector<tree> tr){    int nmax=-1,ge=-1;    int t=1;    if(N==1||N==0)    {        nmax=N;ge=N;        cout<<nmax<<" "<<ge<<endl;        return;    }    int rear=-1,front=-1,level=0,last;    vector<int> thist(N);    thist[++rear]=t;    level=1;    last=rear;    while(front<rear)    {        t=thist[++front];        for(int i=0;i<tr[t]->child.size()&&rear<N;i++)        {            thist[++rear]=tr[t]->child[i];        }        if(last==front)        {            last=rear;            level++;        }        if(rear-front>nmax)        {                        nmax=rear-front;            ge=level;        }    }    cout<<nmax<<" "<<ge<<endl;}