A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.
Input Specification:
Each input file contains one test case. Each case starts with a line containing 0, the number of nodes in a tree, and M (<), the number of non-leaf nodes. Then M lines follow, each in the format:
ID K ID[1] ID[2] ... ID[K]
where ID
is a two-digit number representing a given non-leaf node, K
is the number of its children, followed by a sequence of two-digit ID
's of its children. For the sake of simplicity, let us fix the root ID to be 01
.
The input ends with N being 0. That case must NOT be processed.
Output Specification:
For each test case, you are supposed to count those family members who have no child for every seniority level starting from the root. The numbers must be printed in a line, separated by a space, and there must be no extra space at the end of each line.
The sample case represents a tree with only 2 nodes, where 01
is the root and 02
is its only child. Hence on the root 01
level, there is 0
leaf node; and on the next level, there is 1
leaf node. Then we should output 0 1
in a line.
Sample Input:
2 1
01 1 02
Sample Output:
0 1
#include<bits/stdc++.h> using namespace std; typedef long long ll; #define MAXN 500 vector<vector<int>> vec; int to_int(string sx){ int sum = 0; sum += sx[0]-'0'; sum = sum*10+(sx[1]-'0'); return sum; } int main(){ int n,m; cin >> n >> m; vec.resize(n+1); for(int i=0;i < m;i++){ string s; cin >> s; int x = to_int(s); int K; cin >> K; for(int j=0;j < K;j++){ string temp; cin >> temp; vec[x].push_back(to_int(temp)); } } queue<int> que; que.push(1); while(!que.empty()){ int len = que.size(); int cnt = 0; for(int i=0;i < len;i++){ int t = que.front();que.pop(); if(vec[t].size()){ for(int j=0;j < vec[t].size();j++){ que.push(vec[t][j]); } } else cnt++; } cout << cnt; if(!que.empty()) cout << " "; } return 0; }
这题也30分??
树看成一个有向图,用邻接表储存,然后bfs遍历即可