PAT 1094 The Largest Generation[bfs][一般]
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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
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!
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