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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; }
hash 表 | | jzoj 1335 | | 脑残+手残 | | 集合的关系
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原文地址:http://www.cnblogs.com/supersumax/p/5950219.html