1094 The Largest Generation (25 分)dfs或bfs求图的最大宽度
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
思路:
bfs和dfs都可以,主要是求图的最大宽度,与其对应的层数,dfs更加简洁一些。
#include<iostream> #include<vector> #include<algorithm> #include<queue> #include<string> #include<map> #include<set> using namespace std; vector<vector<int>> pt; vector<int> curLevel; int generation[102]; int visited[101]; void bfs() { queue<int> ids; ids.push(1);//把根的索引入队 visited[1]=1; curLevel[1]=1; generation[1]=1; while(!ids.empty()) { int temp=ids.front(); ids.pop(); for(auto num:pt[temp]) { if(visited[num]==0) { ids.push(num); visited[num]=1; curLevel[num]=curLevel[temp]+1; generation[curLevel[num]]++;//统计当前层的人数 //gr.push_back(gr[temp-1]+1); } } } } //int book[105]; void dfs(int level,int cur) { generation[level]++; for(auto num:pt[cur]) { if(visited[num]==0) { visited[num]=1; dfs(level+1,num,n); } } } int main() { int n,m; cin>>n>>m; pt.resize(n+1); curLevel.resize(n+3); for(int i=0;i<m;i++) { int id0,k; cin>>id0>>k; int temp; for(int j=0;j<k;j++) { cin>>temp; pt[id0].push_back(temp); } } //bfs(); dfs(1,1); int gen=1,level=1; for(int i=1;generation[i]!=0;i++) { if(generation[i]>gen) { gen=generation[i]; level=i; } // cout<<i<<endl; } cout<<gen<<" "<<level; return 0; }
posted on 2019-01-22 18:03 ZhangのBlog 阅读(299) 评论(0) 编辑 收藏 举报