1004. Counting Leaves (30)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.

Input

Each input file contains one test case. Each case starts with a line containing 0 < N < 100, the number of nodes in a tree, and M (< N), 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.

Output

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

解析：通过给你每一个家庭成员的孩子关系，叫你列出每一层的叶子节点的数目，也就是说列出同一辈份没有子女的成员的个数。

Codes:

/*************************************************************************
    > File Name: 1004.cpp
    > Author:
    > Mail:
    > Created Time: 2015年12月08日 星期二 13时14分33秒
 ************************************************************************/

#include<iostream>
#include<queue>
#include<vector>
#include<string>
#include<map>
using namespace std;

struct node{
    string id;
    vector<node*> children;
    node(string x){
        id = x;
    }
};

int main(){
    int n,m;
    cin>>n>>m;
    if(m == 0){
        cout<<n;
        return 0;
    }

    // map store nodes
    map<string, node*> cache;

    string id;
    int k;
    while(m--){
        cin>>id>>k;
        if(cache.count(id) <= 0){
            cache[id] = new node(id);
        }
        string idi;
        int index = 0;
        while(k--){
            cin>>idi;
            node *tmp = NULL;
            if(cache.count(idi) <= 0){
                tmp = new node(idi);
                cache[idi] = tmp;
            }
            else{
                tmp = cache[idi];
            }
            cache[id]->children.push_back(tmp);
            index++;
        }
    }

    string rootid = "01";
    queue<node*> qu;
    qu.push(cache[rootid]);
    while(!qu.empty()){
        int nums = 0;
        int size = qu.size();
        for(int i=0; i<size; i++){
            node *tmp2 = qu.front();
            qu.pop();
            if(tmp2->children.size() == 0){
                nums++;
            }
            else{
                for(int i=0; i<tmp2->children.size(); i++){
                    qu.push(tmp2->children[i]);
                }
            }
        }

        cout<<nums;
        if(!qu.empty()){
            cout<<" ";
        }
    }

    return 0;
}

注：一开始总是有segmentation fault，就仔细得检查，后面才发现原来是没有考虑m=0的情况，也就是说，所有的家庭成员都是同一辈份的人员，所以就只有一层，输出n即可。