hash 表 | | jzoj 1335 | | 脑残+手残 | | 集合的关系

给定两个集合A、B，集合内的任一元素x满足1 ≤ x ≤ 10^9，并且每个集合的元素个数不大于10^5。我们希望求出A、B之间的关系。

给定两个集合的描述，判断它们满足下列关系的哪一种：
A是B的一个真子集，输出“A is a proper subset of B”
B是A的一个真子集，输出“B is a proper subset of A”
A和B是同一个集合，输出“A equals B”
A和B的交集为空，输出“A and B are disjoint”
上述情况都不是，输出“I'm confused!”

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哈希表的线性探测法，比较ab数组项数多少，然后对a（b）数组进行插入，b（a）数组进行查找，flag用于查询即可

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<ctime>
using namespace std;
const int maxprime=1000003;
const int step=7;
const int N=1000000;
struct nums
{
    int val,sum;
}hash[N];
int m,n;
int find(int x)
{
    int temp=x%maxprime;
    while(hash[temp].val!=0&&hash[temp].val!=x)
    {
        temp+=step;
        if(temp>=maxprime)  temp-=maxprime;
    }
    return temp;
}
bool flag1=0,flag2=0;
void  insert(int x)
{
    int now=find(x);    
    hash[now].val=x;
    return ;
}
void judge(int x)
{
    int now=find(x);
    if(hash[now].val==x)    flag1=1;
    else    flag2=1;
}
int a[N],b[N];
int main()
{
    //freopen("a.in","r",stdin);
   //freopen("a.out","w",stdout);
    int i=1;
    ios::sync_with_stdio(false);
    cin>>m;
    for(int i=1;i<=m;i++)
        cin>>a[i];
    cin>>n;
    for(int j=1;j<=n;j++)
        cin>>b[j];
    if(m<n) 
    {
        for(int i=1;i<=n;i++)
           insert(b[i]);
        for(int i=1;i<=m;i++)
            judge(a[i]);
        if(flag1==1&&flag2==0)  {cout<<"A is a proper subset of B"<<endl; /*cout<<"THE TIME HAS PASSED: "<<clock()<<" ms"<<endl;*/return 0;}
        if(flag1==0&&flag2==1)  {cout<<"A and B are disjoint"<<endl; /*cout<<"THE TIME HAS PASSED: "<<clock()<<" ms"<<endl;*/return 0;}
    }
    if(m>n)
    {
        for(int j=1;j<=m;j++)
            insert(a[j]);
        for(int i=1;i<=n;i++)
            judge(b[i]);
        if(flag1==1&&flag2==0)  {cout<<"B is a proper subset of A"<<endl; /*cout<<"THE TIME HAS PASSED: "<<clock()<<" ms"<<endl;*/return 0;}
        if(flag1==0&&flag2==1)  {cout<<"A and B are disjoint"<<endl; /*cout<<"THE TIME HAS PASSED: "<<clock()<<" ms"<<endl;*/return 0;}
    }
    if(m==n)
    {
        for(int i=1;i<=m;i++)
            insert(a[i]);
        for(int j=1;j<=n;j++)
            judge(b[j]);
        if(flag1==1&&flag2==1)  {cout<<"A equals B"<<endl;return 0;}
        if(flag1==0&&flag2==1)  {cout<<"A and B are disjoint"<<endl; /*cout<<"THE TIME HAS PASSED: "<<clock()<<" ms"<<endl;*/return 0;}
    }
    cout<<"I'm confused!"<<endl;
     /*cout<<"THE TIME HAS PASSED: "<<clock()<<" ms"<<endl;*/
    return 0;
   
}