1094 The Largest Generation(25 分)
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
題目比較簡單,求得就是最多孩子的那一層的深度和寬度。
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100 + 10;
int n, m;
struct Node {
int v;
int level;
vector<int> childs;
};
Node people[maxn];
int len[maxn];
void bfs() {
queue<Node> q;
q.push(people[1]);
while (!q.empty()) {
Node f = q.front(); q.pop();
int u = f.v;
for (int i = 0; i < (people[u].childs).size(); i++) {
int v = people[u].childs[i];
people[v].level = f.level + 1;
q.push(people[v]);
}
}
memset(len, 0, sizeof(len));
for (int i = 1; i <= n; i++) {
int level = people[i].level;
len[level]++;
}
int maxlen = 0;
int maxlevel = 0;
for (int i = 0; i < maxn; i++) {
if (maxlen < len[i]) {
maxlen = len[i];
maxlevel = i;
}
}
cout << maxlen << " " << (maxlevel + 1) << endl;
}
int getlen() {
memset(len, 0, sizeof(len));
for (int i = 1; i <= n; i++) {
int level = people[i].level;
len[level]++;
}
int maxlen = 0;
for (int i = 0; i < maxn; i++) {
maxlen = max(maxlen, len[i]);
}
return maxlen;
}
int main() {
scanf("%d %d", &n, &m);
while (m-- != 0) {
int id, num;
scanf("%d", &id);
scanf("%d", &num);
Node cas;
cas.v = id;
cas.level = 0;
cas.childs.clear();
for (int i = 0; i < num; i++) {
int child; cin >> child;
cas.childs.push_back(child);
}
people[id] = cas;
}
bfs();
return 0;
}