ZJCPC 2020 浙江省赛

K. Killing the Brute-force

题目大意

给定两个数组，问右侧的\(a_i*3<b_i\)的位置

解题思路

按题意模拟即可

代码实现

#include <bits/stdc++.h>

using i64 = long long;

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);

    int tt;
    std::cin >> tt;

    while (tt--) {
        int m, ans = -1;
        std::cin >> m;

        std::vector<int> a(m), b(m);
        for (int i = 0; i < m; i++) {
            std::cin >> a[i];
        }
        for (int i = 0; i < m; i++) {
            std::cin >> b[i];
        }
        for (int i = 0; i < m; i++) {
            if (a[i] * 3 < b[i]) {
                ans = i + 1;
                break;
            }
        }

        std::cout << ans << "\n";
    }
}

A. AD 2020

题目大意

给定两个日期，问中间有多少日期包涵“202”

解题思路

预处理后对字典序二分即可

代码实现

#include <bits/stdc++.h>

using i64 = long long;

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);

    std::vector<int> m = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    std::vector<std::string> ans;
    for (int year = 2000; year <= 9999; year++) {
        for (int month = 1; month <= 12; month++) {
            int f = ((year % 4 == 0) && (year % 100 != 0) || (year % 400 == 0)) && (month == 2);
            for (int day = 1; day <= m[month - 1] + f; day++) {
                std::string s = std::to_string(year);
                if (month <= 9) {
                    s += "0";
                }
                s += std::to_string(month);
                if (day <= 9) {
                    s += "0";
                }
                s += std::to_string(day);
                if (s.find("202") != -1) {
                    ans.push_back(s);
                }
            }
        }
    }
    std::sort(ans.begin(), ans.end());

    int tt;
    std::cin >> tt;

    while (tt--) {
        int y1, m1, d1, y2, m2, d2;
        std::cin >> y1 >> m1 >> d1 >> y2 >> m2 >> d2;

        std::string s1 = std::to_string(y1), s2 = std::to_string(y2);
        if (m1 <= 9) {
            s1 += "0";
        }
        s1 += std::to_string(m1);
        if (d1 <= 9) {
            s1 += "0";
        }
        s1 += std::to_string(d1);

        if (m2 <= 9) {
            s2 += "0";
        }
        s2 += std::to_string(m2);
        if (d2 <= 9) {
            s2 += "0";
        }
        s2 += std::to_string(d2);

        std::cout << std::upper_bound(ans.begin(), ans.end(), s2) - std::lower_bound(ans.begin(), ans.end(), s1) << "\n";
    }
}

I. Invoking the Magic

题目大意

给你n对不一定颜色匹配的袜子，将它们重新正确分组，问袜子数量最多的组有多少袜子

解题思路

对袜子颜色编号离散化，再映射颜色和编号并查集维护，答案就是最大的连通块大小

代码实现

#include <bits/stdc++.h>

using i64 = long long;

class DSU {
   public:
    int cnt;
    std::vector<int> fa, rank, siz;

    DSU(int n) : cnt(n), fa(n + 1), rank(n + 1), siz(n + 1, 1) {
        for (int i = 1; i <= n; i++) {
            fa[i] = i;
        }
    }

    int find(int x) {
        if (fa[x] != x) {
            fa[x] = find(fa[x]);
        }
        return fa[x];
    }

    void merge(int x, int y) {
        int X = find(x), Y = find(y);
        if (X != Y) {
            if (rank[X] >= rank[Y]) {
                fa[Y] = X;
                siz[X] += siz[Y];
                if (rank[X] == rank[Y]) {
                    rank[X]++;
                }
            } else {
                fa[X] = Y;
                siz[Y] += siz[X];
            }
            cnt--;
        }
    }

    int size() {
        return cnt;
    }

    int count(int x) {
        return siz[find(x)];
    }
};

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);

    int tt;
    std::cin >> tt;

    while (tt--) {
        int n, ans = 0, cnt = 1;
        std::cin >> n;

        std::set<int> st;
        std::map<int, int> mp;
        std::vector<std::array<int, 2>> g1, g2;
        for (int i = 0; i < n; i++) {
            int u, v;
            std::cin >> u >> v;
            st.insert(u);
            st.insert(v);
            g1.push_back({u, v});
        }

        for (auto x : st) {
            mp[x] = cnt++;
        }

        for (auto [u, v] : g1) {
            g2.push_back({mp[u], mp[v]});
        }

        DSU dsu(n);
        for (auto [u, v] : g2) {
            dsu.merge(u, v);
        }

        for (int i = 1; i <= n; i++) {
            ans = std::max(ans, dsu.count(i));
        }

        std::cout << ans << "\n";
    }
}

C. Crossword Validation

题目大意

给你一个字符矩阵要求计算所有匹配串的权值之和，无法匹配输出-1

解题思路

trie来记录m个单词以及它们的权值，之后遍历矩阵查找。（直接umap好像也跑的飞快）

代码实现

#include <bits/stdc++.h>

using i64 = long long;

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);

    int tt;
    std::cin >> tt;

    while (tt--) {
        int n, m;
        std::cin >> n >> m;

        std::vector<std::string> g(n), S;
        for (int i = 0; i < n; i++) {
            std::cin >> g[i];
            std::string s;
            for (auto ch : g[i]) {
                if (ch != '#') {
                    s += ch;
                } else {
                    if (s.size()) {
                        S.push_back(s);
                    }
                    s.clear();
                }
            }
            if (s.size()) {
                S.push_back(s);
            }
        }

        for (int j = 0; j < n; j++) {
            std::string s;
            for (int i = 0; i < n; i++) {
                if (g[i][j] != '#') {
                    s += g[i][j];
                } else {
                    if (s.size()) {
                        S.push_back(s);
                    }
                    s.clear();
                }
            }
            if (s.size()) {
                S.push_back(s);
            }
        }

        std::unordered_map<std::string, int> ump;
        for (int i = 0; i < m; i++) {
            int x;
            std::string s;
            std::cin >> s >> x;
            ump[s] = x;
        }

        i64 ans = 0;
        for (auto s : S) {
            if (ump.count(s)) {
                ans += ump[s];
            } else {
                ans = -1;
                break;
            }
        }

        std::cout << ans << "\n";
    }
}