pat甲 1004 Counting Leaves (30)

題目大意：N代表N個節點。M代表，接下來M行輸入。

每一行：ID代表父節點，k代表有幾個兒子，ID[x]代表兒子是那些。

輸出關係樹每一層的沒有兒子的節點數，也就是葉子數。

思路：把關係存儲起來，找到父節點

遍歷找每一層葉子數，用數組存儲起來

輸出

 

 

#include<iostream>
#include<vector>
#include<map>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn = 105;
vector<int> v[maxn];
map<string, int> mp;
int counts, n, m, maxLevel;
int ans[maxn];
int get_id(string n)
{
    if(!mp.count(n))
        mp[n] = counts++;
    return mp[n];
}
int findFather()
{
    //cout<<"yes"<<endl;
    int vis[n];
    memset(vis, 0, sizeof(vis));
    for(int i = 0; i < n; i++)
        for(int j = 0; j < v[i].size(); j++)
        {
            //cout<<v[i][j]<<endl;
            vis[v[i][j]] = 1;
        }
    for(int i = 0; i < n; i++)
    {
        if(!vis[i])
            return i;
    }
}
void travel(int root, int level)
{
    maxLevel = max(level, maxLevel);
    if(!v[root].size()) ans[level]++;
    for(int i = 0; i < v[root].size(); i++)
    {
        int son = v[root][i];
        travel(son, level + 1);
    }
}
void solve()
{
    memset(ans, 0, sizeof(ans));
    int f = findFather();
    //cout<<f<<endl;
    travel(f, 0);
    for(int i = 0; i <= maxLevel; i++)
    {
        cout<<ans[i];
        if(i != maxLevel) cout<<" ";
    }
    cout<<endl;

}
int main()
{
    while(cin>>n>>m)
    {
        maxLevel = 0;
        counts = 0;
        mp.clear();
        for(int i = 0; i <= n; i++)
            v[i].clear();
        for(int i = 0; i < m; i++)
        {
            string id;
            int k;
            cin>>id>>k;
            int newId = get_id(id);
            for(int j = 0; j < k; j++)
            {
                string sons;
                cin>>sons;
                int sonsId = get_id(sons);
                v[newId].push_back(sonsId);
            }
        }
        solve();

    }


    return 0;
}