PAT (Advanced Level) Practise 1094 The Largest Generation (25)

最新推荐文章于 2019-03-27 23:19:34 发布

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最新推荐文章于 2019-03-27 23:19:34 发布

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本文介绍了一道关于寻找家族树中拥有最多成员的一代的算法题目。通过深度优先搜索（DFS）的方法遍历树状结构，统计各代的人口数量，并最终找出人数最多的那一代及其具体人数。

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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

题意：给你一棵有n个节点的树，问那一层的节点最多，有几个

解题思路：dfs遍历一遍树

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <cmath>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

#define mod 1000000007

int n,m,ma;
int level[105];
int s[105],nt[105],e[105];

void dfs(int k,int p)
{
    level[p]++;
    if(level[p]>level[ma]) ma=level[p];
    for(int i=s[k];~i;i=nt[i])
    {
        int ee=e[i];
        dfs(ee,p+1);
    }
}

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        int id,k,cnt=1,p;
        memset(level,0,sizeof level);
        memset(s,-1,sizeof s);
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d",&id,&k);
            for(int j=1;j<=k;j++)
            {
                scanf("%d",&p);
                nt[cnt]=s[id],s[id]=cnt,e[cnt++]=p;
            }
        }
        ma=-1;
        dfs(1,1);
        printf("%d %d\n",level[ma],ma);
    }
    return 0;
}