2.6

PTA学习数据结构之旅

树的同构

https://pintia.cn/problem-sets/15/problems/711

解法①

#include<stdio.h>
#define max 12
#define Null -1
struct node
{
    char data;
    int left,right;
}t1[12],t2[12];
int build(struct node t[],int n)
{    
    int i,check[12]={0},root=Null;
    if(n)
    {
        
        for(i=0;i<n;i++)
        {
            char c1,c2;
            scanf("\n%c %c %c",&t[i].data,&c1,&c2);
            if(c1!='-')
            {
                t[i].left=c1-48;
                check[t[i].left]=1;
            }
            else t[i].left=-1;
            if(c2!='-')
            {
                t[i].right=c2-48;
                check[t[i].right]=1;
            }
            else t[i].right=-1; 
        }
        for(i=0;i<n;i++)
            if(!check[i]) break;
        root=i;
    }
    return root;
}
int is(int r1,int r2)
{
    if(r1==-1&&r2==-1)return 1;
    if (r1==-1&&r2!=-1||r1!=-1&&r2==-1) return 0;
    if(t1[r1].data!=t2[r2].data) return 0;
    if(t1[r1].left==-1&&t2[r2].left==-1)return is(t1[r1].right,t2[r2].right);
    if(t1[r1].right==-1&&t2[r2].right==-1)return is(t1[r1].left,t2[r2].left);
    if(t1[r1].left!=-1&&t2[r2].left!=-1&&t1[t1[r1].left].data==t2[t2[r2].left].data)
        return (is(t1[r1].left,t2[r2].left))&&(is(t1[r1].right,t2[r2].right));
    else return (is(t1[r1].right,t2[r2].left))&&(is(t1[r1].left,t2[r2].right));
    
}
int main()
{    
    int r1,r2,n1,n2;
    scanf("%d",&n1);
    r1=build(t1,n1);
    scanf("%d",&n2);
    r2=build(t2,n2);
    printf("%s",is(r1,r2)?"Yes":"No");
    
}

解法②运用向量

题中说树的同构就是左右孩子交换，那么左右孩子的ASCII值之和就是不变的，是一直相等的。

#include<iostream>
#include<vector>
#include<cctype>
using namespace std;
struct node
{
    char a;
    char b;
    char c;
};
int main()
{
    int n1, n2;
    cin >> n1;
    vector<node>v1(n1);
    for (int i = 0; i < n1; i++)
    {
        cin >> v1[i].a >> v1[i].b >> v1[i].c;
    }
    cin >> n2;
    vector<node>v2(n2);
    for (int i = 0; i < n2; i++)
    {
        cin >> v2[i].a >> v2[i].b >> v2[i].c;
    }
    if (n1 == 1 && n2 == 1)//对一个节点的特判
    {
        cout << "No";
        return 0;
    }
    for (int i = 0; i < v1.size(); i++)
    {
        int value1 = 0, value2 = 0;
        for (int j = 0; j < v2.size(); j++)
        {
            if (v1[i].a == v2[j].a)
            {
                if (isdigit(v1[i].b)) value1 += v1[v1[i].b - '0'].a;
                if (isdigit(v1[i].c)) value1 += v1[v1[i].c - '0'].a;
                if (isdigit(v2[j].b)) value2 += v2[v2[j].b - '0'].a;
                if (isdigit(v2[j].c)) value2 += v2[v2[j].c - '0'].a;
                if (value1 != value2)
                {
                    cout << "No";
                    return 0;
                }
            }

        }

    }
    cout << "Yes";
    return 0;
}