第9章 哈希表

1048. Find Coins

笔记

• 哈希表存储
• 每读入一个数x时，在哈希表中查找target - x是否存在，如果存在则检查是否需要更新全局最优解

#include <iostream>
#include <unordered_set>
using namespace std;
const int INF = 1010;
unordered_set<int> h;

int main() {
    int n, target;
    cin >> n >> target;

    int x, p = INF, q = INF;
    while(n--) {
        cin >> x;
        int y = target - x;
        if (h.count(y)) {
            if (x > y) swap(x, y);
            if (x < p) {
                p = x;
                q = y;
            }
        }
        h.insert(x);
    }

    if (p == INF) puts("No Solution");
    else cout << p << ' ' << q << endl;

    return 0;
}

1063. Set Similarity

笔记

• 用哈希表存储集合$A$和$B$
• 在求集合$A \cap B$的元素个数时，可先遍历集合$A$的元素，如果该元素在集合$B$中存在，则计数+1，时间复杂度是$O(n+m)$，而$A \cup B$的元素个数时可根据公式在$O(1)$时间复杂度计算出

#include <iostream>
#include <unordered_set>
using namespace std;
const int N = 52;
unordered_set<int> s[N];

int main() {
    int n, m, x;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &m);
        while(m--) {
            scanf("%d", &x);
            s[i].insert(x);
        }
    }

    int q, a, b;
    scanf("%d", &q);
    while(q--) {
        scanf("%d%d", &a, &b);

        int cnt = 0;
        for (int x : s[a])
            if (s[b].count(x))
                cnt++;

        double res = cnt * 100.0 / (s[a].size() + s[b].size() - cnt);
        printf("%.1lf%%\n", res);
    }

    return 0;
}

1120. Friend Numbers

笔记

• 用集合set存储就行，不用统计有多少次
• 遍历set时，默认是升序遍历

#include <iostream>
#include <set>
using namespace std;
set<int> s;


int main() {
    int n, x;
    cin >> n;
    while(n--) {
        cin >> x;
        int sum = 0;
        while(x) {
            sum += x % 10;
            x /= 10;
        }
        s.insert(sum);
    }

    cout << s.size() << endl;
    string res;
    for (auto elem : s)
        res += to_string(elem) + ' ';
    res.pop_back();
    cout << res << endl;

    return 0;
}

1144. The Missing Number

笔记

• 插入到哈希表，然后再依次查询

#include <iostream>
#include <unordered_set>
using namespace std;
unordered_set<int> s;

int main() {
    int n, x;
    cin >> n;
    while(n--) {
        cin >> x;
        s.insert(x);
    }

    int k = 1;
    while(true) {
        if (!s.count(k)) {
            cout << k << endl;
            break;
        }
        k++;
    }

    return 0;   
}

笔记

#include <iostream>
#include <unordered_set>
using namespace std;
const int N = 1e4 + 10;
int a[N][2];

int main() {
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < n; i++) 
        cin >> a[i][0] >> a[i][1];

    int x;
    unordered_set<int> s;
    while(m--) {
        s.clear();
        int t;
        cin >> t;
        while(t--) {
            cin >> x;
            s.insert(x);
        }

        bool success = true;
        for (int i = 0; i < n; i++)
            if (s.count(a[i][0]) && s.count(a[i][1])) {
                success = false;
                break;
            }

        if (success) puts("Yes");
        else puts("No");
    }

    return 0;
}

1149. Dangerous Goods Packaging

笔记

• 先存储所有危险的组合，然后读入集装箱的物品id，遍历所有危险组合，如果每一种组合的两个元素同时再集装箱中出现，则危险

#include <iostream>
#include <unordered_set>
using namespace std;
const int N = 1e4 + 10;
int a[N][2];

int main() {
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < n; i++) 
        cin >> a[i][0] >> a[i][1];

    int x;
    unordered_set<int> s;
    while(m--) {
        s.clear();
        int t;
        cin >> t;
        while(t--) {
            cin >> x;
            s.insert(x);
        }

        bool success = true;
        for (int i = 0; i < n; i++)
            if (s.count(a[i][0]) && s.count(a[i][1])) {
                success = false;
                break;
            }

        if (success) puts("Yes");
        else puts("No");
    }

    return 0;
}

1078. Hashing

笔记

• 模拟二次探测法（把哈希表看做循环数组）
• 注意把长度不是质数的哈希表长度变成质数长度

#include <iostream>
using namespace std;
const int N = 1e4 + 10;
int m, n, h[N];

bool is_prime(int x) {
    if (x < 2) return false;

    for (int i = 2; i <= x / i; i++)
        if (x % i == 0)
            return false;
    return true;
}

int get_pos(int x) {
    int t = x % m;
    for (int k = 0; k < m; k++) {
        int p = (t + k * k) % m;        // 循环数组
        if (!h[p]) return p;
    }
    return -1;      // 无法插入
}

int main() {
    cin >> m >> n;

    while (!is_prime(m)) m++;       // 修正成质数

    int x;
    string res;
    while(n--) {
        cin >> x;
        int t = get_pos(x);

        if (t == -1) res += "- ";
        else {
            h[t] = x;
            res += to_string(t) + ' ';
        }
    }
    res.pop_back();
    cout << res << endl;

    return 0;
}

1137. Final Grading

笔记

• 学生信息是结构体，但由于名字是字符串，用哈希表unordered_map<string, Student>可以加快查找效率

#include <iostream>
#include <cstring>
#include <unordered_map>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;
const int N = 1e4 + 10, M = 21;
char id[M];

struct Student {
    string id;
    int gp, gm, gf, g;

    Student(): gp(-1), gm(-1), gf(-1), g(0) {};

    bool operator< (const Student& s) const {
        if (g != s.g) return g > s.g;
        else return id < s.id;
    }

};

unordered_map<string, Student> students;

int main() {
    int p, m, n;
    scanf("%d%d%d", &p, &m, &n);

    int score;
    for (int i = 0; i < p; i++) {
        scanf("%s%d", id, &score);
        students[id].gp = score;
        students[id].id = id;
    }
    for (int i = 0; i < m; i++) {
        scanf("%s%d", id, &score);
        students[id].gm = score;
        students[id].id = id;
    }
    for (int i = 0; i < n; i++) {
        scanf("%s%d", id, &score);
        students[id].gf = score;
        students[id].id = id;
    }

    vector<Student> res;
    for (auto item : students) {
        Student stu = item.second;
        if (stu.gf > stu.gm) stu.g = stu.gf;
        else stu.g = round(stu.gf * 0.6 + stu.gm * 0.4);
        if (stu.gp >= 200 && stu.g >= 60)
            res.push_back(stu);
    }

    sort(res.begin(), res.end());

    for (auto s : res)
        cout << s.id << ' ' << s.gp << ' ' << s.gm << ' ' << s.gf << ' ' << s.g << endl;

    return 0;
}

1145. Hashing - Average Search Time

笔记

• 模拟哈希表，采用二次探测法解决冲突，需要计算平均查询次数
  • 当查到该元素或发现可用的空位时，查询次数等于进入循环的次数
  • 当查不到该元素时，查询次数等于哈希表容量+1，依旧等价于进入循环的次数

#include <iostream>
using namespace std;
const int N = 1e4 + 10;
int m, n, q, h[N];

bool is_prime(int x) {
    if (x < 2) return false;

    for (int i = 2; i <= x / i; i++)
        if (x % i == 0)
            return false;

    return true;
}

int get_pos(int x, int& cnt) {
    cnt = 1;            // 进入循环前就认为查询一次
    int t = x % m;
    for (int k = 0; k < m; k++) {
        int p = (t + k * k) % m;
        if (!h[p] || h[p] == x) 
            return p;
        cnt++;
    }
    return -1;          // 全都查不到时，返回cnt == m + 1
}

int main() {
    cin >> m >> n >> q;

    while(!is_prime(m)) m++;

    int cnt, t, x;
    for (int i = 0; i < n; i++) {
        cin >> x;
        t = get_pos(x, cnt);
        if (t == -1) printf("%d cannot be inserted.\n", x);
        else h[t] = x;
    }

    int sum = 0;
    for (int i = 0; i < q; i++) {
        cin >> x;
        get_pos(x, cnt);
        sum += cnt;
    }

    printf("%.1lf\n", (double) sum / q);

    return 0;
}