順序表實(shí)現(xiàn)集合的交叉并集運(yùn)算

1、順序表實(shí)現(xiàn)集合的交叉并集運(yùn)算 #include<iostream>#include<stdlib.h>using namespace std;#define maxsize 50typedef structdouble datamaxsize;int last;seqlist;seqlist *init_seqlist() /為順序表分配存儲空間 seqlist *l;l=new seqlist;if(l)l->last=-1;return l;elsel->last=0;return l;void creat

2、e_seqlist(seqlist *L) /初始化集合中的元素 int length;cout<<"請輸入集合中元素的個數(shù)"<<endl;cin>>length;cout<<"請輸入集合中的元素"<<endl;for(int i=0;i<length;i+)cin>>L->datai;L->last+;void p

3、rint(seqlist *L) /輸出集合中的元素for(int i=0;i<=L->last;i+)cout<<L->datai<<" "cout<<"n"void jiaoji(seqlist *L1,seqlist *L2) /求兩個集合的交集double amaxsize;int b=0;for(int i=0;i<=L1->last;i+)for(int j=

4、0;j<=L2->last;j+)if(L1->datai=L2->dataj)ab=L1->datai; b+;if(i=L1->last)cout<<"兩個集合的交集為"<<endl;for(i=0;i<b;i+)cout<<ai<<" "cout<<"n&q

5、uot;break;void chaji(seqlist *L1,seqlist *L2) /求兩個集合的差集double amaxsize; /存交集int b=0;int i,j,w;for(i=0;i<=L1->last;i+)for(j=0;j<=L2->last;j+)if(L1->datai=L2->dataj)ab=L1->datai; b+;double cmaxsize; /存并集int d=0;for(i=0;i<=L1->last;i+)ci=L1

6、->datai;d+;for(j=0;j<=L2->last;j+)cd+j=L2->dataj; d=L1->last+L2->last+1;for(i=0;i<=d;i+)for(j=i+1;j<=d;j+)if(ci=cj)for(w=j;w<d;w+)cw=cw+1;d-;for(i=0;i<=d;i+)for(j=0;j<b;j+)if(ci=aj)if(i=d)d-;elsefor(w=i;w<d;w+)cw=cw+

7、1;d-;cout<<"n兩個集合的差集為"<<endl;for(j=0;j<=d;j+) cout<<cj<<" "cout<<"n"void bingji(seqlist *L1,seqlist *L2) /求兩個集合的并集double amaxsize;int b=0;for(int i=0;i<=L1-&g

8、t;last;i+)ai=L1->datai;b+;for(int j=0;j<=L2->last;j+)ab+j=L2->dataj; b=L1->last+L2->last+1;for(i=0;i<=b;i+)for(int j=i+1;j<=b;j+)if(ai=aj)for(int w=j;w<b;w+)aw=aw+1;b-;cout<<"兩個集合的并集為"<<endl;

9、for(j=0;j<=b;j+) cout<<aj<<" "cout<<"n"void main()int i;seqlist *L1;L1=init_seqlist();seqlist *L2;L2=init_seqlist();AA:cout<<"* *n"cout<<"* 按鈕1 創(chuàng)建集合 *n&quot

10、;cout<<"* 按鈕2 求集合交集 *n"cout<<"* 按鈕3 求集合并集 *n"cout<<"* 按鈕4 求集合差集 *n"cout<<"* 按鈕5 顯示集合 *n"cout<<"* 按鈕6 退出 *n"cout<<"*n&

11、amp;quot;cout<<"請根據(jù)所需要的功能選擇相應(yīng)的按鈕n"cin>>i;switch(i)case 1:create_seqlist(L1);create_seqlist(L2);goto AA;case 2:jiaoji(L1,L2); goto AA;case 3: bingji(L1,L2);goto AA;case 4: chaji(L1,L2); goto AA;case 5: cout<<"第一個集合中的元素為n"print(L1);cout<<&a