标签：ret long 题目 otto http printf archive hiera text
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.
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.
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.
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
9 4题目大意：求树的具有最多节点的层数， 并输出是第几层。
这是我写的。
#include <iostream> #include <vector> #include <map> #include<stdio.h> #include<cmath> #include<queue> using namespace std; vector<int> tree[202]; map<int,int> mp; int main() { int n,m; queue<int> que; scanf("%d %d",&n,&m); int id,k,temp; for(int i=0;i<m;i++){ scanf("%d %d",&id,&k); for(int j=0;j<k;j++){ scanf("%d",&temp); tree[id].push_back(temp); } } que.push(1); que.push(-1); mp[1]=1; int level=2; while(!que.empty()){ int top=que.front();que.pop(); while(top!=-1){ for(int i=0;i<tree[top].size();i++){ mp[level]++; que.push(tree[top][i]); } top=que.front();que.pop(); } //是不是得两层while,每次写这个地方都有疑惑的。 if(!que.empty()){ level++;que.push(-1); } } int maxs=0; id=0; for(int i=1;i<=level;i++){ if(mp[i]>maxs){ maxs=mp[i]; id=i; } } printf("%d %d",maxs,id); return 0; }
//用-1作为标记，双层while循环进行遍历。
下列代码均来自：https://www.liuchuo.net/archives/2223
#include <cstdio> #include <vector> using namespace std; vector<int> v[100]; int book[100]; void dfs(int index, int level) { //传参是本节点下标和节点所在层数！ book[level]++; for(int i = 0; i < v[index].size(); i++) dfs(v[index][i], level+1); } int main() { int n, m, a, k, c; scanf("%d %d", &n, &m); for(int i = 0; i < m; i++) { scanf("%d %d",&a, &k); for(int j = 0; j < k; j++) { scanf("%d", &c); v[a].push_back(c); } } dfs(1, 1); int maxnum = 0, maxlevel = 1; for(int i = 1; i < 100; i++) { if(book[i] > maxnum) { maxnum = book[i]; maxlevel = i; } } printf("%d %d", maxnum, maxlevel); return 0; }
//没看到代码之前我都不知道还可以用dfs写。以为只可以用bfs，学习了！
PAT 1094 The Largest Generation[bfs][一般]
标签：ret long 题目 otto http printf archive hiera text
原文地址：https://www.cnblogs.com/BlueBlueSea/p/9517218.html
