PAT 1004 Counting Leaves

1004 Counting Leaves (30 分)

 

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