PAT Advanced Level 1094

1094 The Largest Generation（25 分）

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
没难度 一次过 一看AC率高达0.57 大家都这么棒棒啊

/**********************
author: yomi
date: 18.8.20
ps:
**********************/
#include <iostream>
#include <vector>
#include <cstring>
using namespace std;

struct Node
{
    vector<int>child;
    int id;
}node[1010];
bool vis[1010];
int depth = 1, ans[1010];

void dfs(int index)
{
    if(node[index].child.size() == 0){
        return;
    }
    vis[index] = true;
    for(int i=0; i<node[index].child.size(); i++){
        if(!vis[node[index].child[i]]){
            ans[depth]++;
            depth++;
            dfs(node[index].child[i]);
            depth--;
        }
    }


}
int main()
{
    int n, m, id, t, d;
    memset(ans, 0, sizeof(ans));
    cin >> n >> m;
    for(int i=0; i<m; i++){
        cin >> id >> t;
        for(int j=0; j<t; j++){
            cin >> d;
            node[id].child.push_back(d);
        }
    }
    dfs(1);
    int flag = 1;
    int Max = 0, ind;
    for(int i=1; i<1010; i++){
        if(ans[i] == 0)
            break;
        if(ans[i] > Max){
            Max = ans[i];
            flag = 0;
            ind = i;
        }
    }
    if(flag == 0)
        cout << Max << ' ' << ind+1;
    else{
        cout << "1 1";
    }
    return 0;
}
/**
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
**/

/**********************
author: yomi
date: 18.8.20
ps:
**********************/
#include <iostream>
#include <vector>
#include <cstring>
using namespace std;

struct Node
{
    vector<int>child;
    int id;
}node[1010];
bool vis[1010];
int depth = 1, ans[1010];

void dfs(int index)
{
    if(node[index].child.size() == 0){
        return;
    }
    vis[index] = true;
    for(int i=0; i<node[index].child.size(); i++){
        if(!vis[node[index].child[i]]){
            ans[depth]++;
            depth++;
            dfs(node[index].child[i]);
            depth--;
        }
    }

}
int main()
{
    int n, m, id, t, d;
    memset(ans, 0, sizeof(ans));
    cin >> n >> m;
    for(int i=0; i<m; i++){
        cin >> id >> t;
        for(int j=0; j<t; j++){
            cin >> d;
            node[id].child.push_back(d);
        }
    }
    dfs(1);
    int flag = 1;
    int Max = 0, ind;
    for(int i=1; i<1010; i++){
        if(ans[i] == 0)
            break;
        if(ans[i] > Max){
            Max = ans[i];
            flag = 0;
            ind = i;
        }
    }
    if(flag == 0)
        cout << Max << ' ' << ind+1;
    else{
        cout << "1 1";
    }
    return 0;
}
/**
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
**/