数据结构 求集合的交并补集 (STL)

Description

任意给定两个包含1-30000个元素的集合A,B（集合中元素类型为任意整型数，且严格递增排列），求A交B、A并B、A-B和B-A集合。

Input

输入第一行为测试数据组数。每组测试数据两行，分别为集合A、B。每行第一个数n（1<=n<=30000）为元素数量，后面有n个严格递增的绝对值小于2^31代表集合中包含的数。

Output

对每组测试数据输出5行，第1行为数据组数，后4行分别为按升序输出两个集合的A交B、A并B、A-B和B-A集合。格式见样例。

Sample Input

1

3 1 2 5

4 2 3 5 8

Sample Output

Case #1:

2 5

1 2 3 5 8

1

3 8

HINT

考察知识点：有序表合并，时间复杂度O(n)，空间复杂度O(n)

Append Code

析：这个题用set模拟就好，反正set都有这些函数。也可以自己写。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define frer freopen("in.txt", "r", stdin)
#define frew freopen("out.txt", "w", stdout)
using namespace std;
 
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
set<int> A, B, C;
set<int> :: iterator it;
 
void print(set<int> &C){
    for(it = C.begin(); it != C.end(); ++it){
        if(it == C.begin())  printf("%d", *it);
        else printf(" %d", *it);
    }
    printf("
");
}
 
int main(){
    int T;  cin >> T;
    for(int kase = 1; kase <= T; ++kase){
        printf("Case #%d:
", kase);
        A.clear();  B.clear();
        scanf("%d", &n);
        for(int i = 0; i < n; ++i){
            scanf("%d", &m);
            A.insert(m);
        }
        scanf("%d", &n);
        for(int i = 0; i < n; ++i){
            scanf("%d", &m);
            B.insert(m);
        }
        C.clear();
        set_intersection(ALL(A), ALL(B), INS(C));
        print(C);
        C.clear();
        set_union(ALL(A), ALL(B), INS(C));
        print(C);
        C.clear();
        set_difference(ALL(A), ALL(B), INS(C));
        print(C);
        C.clear();
        set_difference(ALL(B), ALL(A), INS(C));
        print(C);
    }
    return 0;
}