有根树的表达

2019-08-02

11:17:53

题目链接

https://vjudge.net/problem/Aizu-ALDS1_7_A

#include<bits/stdc++.h>
using namespace std;
#define MAX 100005
#define NIL -1
struct Node{
    int p; //节点U的父节点 
    int l; //最左侧的子节点 
    int r; //右侧紧邻的兄弟节点 
}T[MAX];
int n;
int D[MAX];

void print(int u)
{
    cout << "node " << u <<": ";
    cout << "parent = " << T[u].p <<", ";
    cout << "depth = " << D[u] << ", ";
    
    if(T[u].p == NIL) cout << "root, ";
    else if(T[u].l == NIL) cout << "leaf, ";
    else cout << "internal node, ";
    
    cout << "[";
    for(int i = 0, c = T[u].l; c != NIL; i++, c = T[c].r)
    {
        if(i) cout << ", ";
        cout << c;
    }
    cout << "]" << endl;
}
int rec(int u,int p) // 根节点，p是深度 
{
    D[u] = p; //这个是为了记录每一个节点的深度 
    if(T[u].r != NIL) rec(T[u].r,p);
    if(T[u].l != NIL) rec(T[u].l, p+1);
} 
int main()

{
    cin >> n;
    int id, k, c, l, r;
    for(int i = 0; i < n; i++) //初始化 
    {
        T[i].p = T[i].l = T[i].r = NIL; 
    }
    for(int i = 0; i < n; i++)
    {
        cin >> id >> k ; //id节点的编号，k节点的度数    
        for(int j = 0; j < k; j++)
        {
            cin >> c;
            if(j == 0)
            {
                T[id].l = c; // 编号为id的最左侧子节点 
            }
            else
            {
                T[l].r = c;
            }
            l = c;
            T[c].p = id; //存每个节点的父节点 
        }
    }
    
    for(int i = 0; i < n; i++)
    {
        if(T[i].p == NIL) r = i;
    }
    
    rec(r, 0);
    for(int i = 0; i < n; i++)
    {
        print(i); //print是一个函数 
    }
    return 0;    
}