AtCoder Beginner Contest 383

//前三题都很水，只能写写这种题骗自己了

A - Humidifier 1

直接模拟

#include <bits/stdc++.h>

using namespace std;
#define int long long
#define inf INT32_MAX
#define PII pair<int,int>
#define endl '\n'

inline void solve() {
    int n;
    cin >> n;
    int ans = 0;
    int t = 0;
    for (int i = 1; i <= n; i++) {
        int x, y;
        cin >> x >> y;
        if (i == 1) {
            t = x;
            ans = y;
        } else {
            ans = max(ans - (x - t), 0ll);
            ans += y;
            t = x;
        }
    }
    cout << ans << endl;
}


signed main() {
#ifdef ONLINE_JUDGE
#else
    freopen("test.in", "r", stdin);
    freopen("test.out", "w", stdout);
#endif
    ios::sync_with_stdio(0);
    cout.tie(0);
    cin.tie(0);
    int t = 1;
//    cin >> t;
    while (t--)
        solve();
    return 0;
}

B - Humidifier 2

暴力枚举即可

#include <bits/stdc++.h>

using namespace std;
#define int long long
#define inf INT32_MAX
#define PII pair<int,int>
#define endl '\n'

inline void solve() {
    int n, m, k;
    cin >> n >> m >> k;
    vector<PII > num;
    vector<vector<char>> arr(n + 2, vector<char>(m + 2));
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            cin >> arr[i][j];
            if (arr[i][j] == '.')num.push_back(make_pair(i, j));
        }
    }
    int ans = 0;
    auto get_sum = [&](int a, int b, int c, int d) {
        int sum = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                if (arr[i][j] == '.') {
                    int dis1 = abs(a - i) + abs(b - j);
                    int dis2 = abs(c - i) + abs(d - j);
                    if (min(dis1, dis2) <= k)sum++;
                }
            }
        }
        return sum;
    };
    for (int i = 0; i < num.size(); i++) {
        for (int j = i + 1; j < num.size(); j++) {
            ans = max(get_sum(num[i].first, num[i].second, num[j].first, num[j].second), ans);
        }
    }
    cout << ans << endl;
}


signed main() {
#ifdef ONLINE_JUDGE
#else
    freopen("test.in", "r", stdin);
    freopen("test.out", "w", stdout);
#endif
    ios::sync_with_stdio(0);
    cout.tie(0);
    cin.tie(0);
    int t = 1;
//    cin >> t;
    while (t--)
        solve();
    return 0;
}

C - Humidifier 3

直接暴力 $b f s$

#include <bits/stdc++.h>

using namespace std;
#define int long long
#define inf INT32_MAX
#define PII pair<int,int>
#define endl '\n'
int dx[4] = {1, -1, 0, 0};
int dy[4] = {0, 0, 1, -1};
struct Node {
    int x, y, s;
};

inline void solve() {
    int n, m, k;
    cin >> n >> m >> k;
    int ans = 0;
    queue<Node> q;
    vector<vector<bool>> st(n + 1, vector<bool>(m + 1));
    vector<vector<char>> arr(n + 1, vector<char>(m + 1));
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            cin >> arr[i][j];
            if (arr[i][j] == 'H') {
                q.push({i, j, k - 1});
                st[i][j] = true;
                ans++;
            }
        }
    }
    while (q.size()) {
        for (int i = 0; i < 4; i++) {
            int x = q.front().x + dx[i];
            int y = q.front().y + dy[i];
            if (x < 1 || x > n || y < 1 || y > m)continue;
            if (arr[x][y] == '#' || st[x][y] || q.front().s < 0)continue;
            if (q.front().s)q.push({x, y, q.front().s - 1});
            st[x][y] = true;
            ans++;
        }
        q.pop();
    }
    cout << ans << endl;
}

signed main() {
#ifdef ONLINE_JUDGE
#else
    freopen("test.in", "r", stdin);
    freopen("test.out", "w", stdout);
#endif
    ios::sync_with_stdio(0);
    cout.tie(0);
    cin.tie(0);
    int t = 1;
//    cin >> t;
    while (t--)
        solve();
    return 0;
}

D - 9 Divisors

假设一个数分解质因数结果为

$x = x^{p_{1}} \cdot x^{p_{2}} \cdot \cdot \cdot \cdot \cdot x^{p_{n}}$

那么x的因子数等于 $n u m = (p_{1} + 1) \cdot (p_{2} + 1) \cdot \cdot \cdot \cdot \cdot (p_{n} + 1)$

由此可知让因子数为9只有两种情况一种情况是 $p_{1} = 2, p_{2} = 2$ 另一种情况是 $p_{1} = 8$

对于第一种情况需要确保 ${(p_{1} \cdot p_{2})}^{2} \leq N$ ，显然对于另一种情况 $p_{1}^{8} \leq N$ 由此可以计算出答案

#include <bits/stdc++.h>

using namespace std;
#define int long long
#define inf INT32_MAX
#define PII pair<int,int>
#define endl '\n'
const int N = 2e6 + 7;
int prime[N];//质数存储于prime数组中，并且从下标0开始存储
bool vis[N];

int euler_sieve(int n) {
    int cnt = 0;
    memset(prime, 0, sizeof prime);
    memset(vis, false, sizeof vis);
    for (int i = 2; i <= n; i++) {
        if (!vis[i])prime[cnt++] = i;
        for (int j = 0; j < cnt; j++) {
            if (i * prime[j] > n)break;
            vis[i * prime[j]] = true;
            if (i * prime[j] == 0)break;
        }
    }
    return cnt;
}

inline void solve() {
    int n;
    cin >> n;
    int ans = 0;
    int cnt = euler_sieve(sqrt(n));
    prime[cnt] = inf;
    for (int i = 0; i < cnt; i++) {
        int tmp = sqrt((n / pow(prime[i], 2)));
        int pos = upper_bound(prime, prime + cnt, tmp) - prime - 1;
        if (pos <= i)break;
        ans += pos - i;
    }
    for (int i = 0;; i++) {
        if (pow(prime[i], 8) <= n)ans++;
        else break;
    }
    cout << ans << endl;
}


signed main() {
#ifdef ONLINE_JUDGE
#else
    freopen("test.in", "r", stdin);
    freopen("test.out", "w", stdout);
#endif
    ios::sync_with_stdio(0);
    cout.tie(0);
    cin.tie(0);
    int t = 1;
//    cin >> t;
    while (t--)
        solve();
    return 0;
}

E - Sum of Max Matching

//挺好的一道题

观察题目，可以交换数组B的位置其实等价于可以可以交换A的位置，因为只需要最后答案，当然知道了这个也没有什么用。要让所有路径上最大值之和最小，如何确保最小呢？联想到最小生成树思考 $K r u s k a l$ 每次将最小权值的边放入图中，如果两点已经相连则不需要。

思考如果A集合中与B集合由于一次 $K r u s k a l$ 加边导致相连那么此时是不是一定是，两点相连的所有可能路径中最大值最小的一条？由于是升序加边确保了每次加边的权值都是当前最小的，那么第一次让两点相连的时候一定是路径上最大值最小的时候。

现在面对第二个问题，如何维护两个集合之间的相连情况呢？记录A与B数组中每个数出现次数，每次添边的时候新增的答案数量就是两者最小值，然后将剩余的边传递到该节点祖先以供后续使用

#include <bits/stdc++.h>

using namespace std;
#define int long long
#define inf INT64_MAX
#define PII pair<int,int>
#define endl '\n'
const int N = 2e5 + 7;
int fa[N];

struct Node {
    int u, v, w;

    Node() {}

    Node(int u, int v, int w) : u(u), v(v), w(w) {}
} node[N * 2];

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

void merge(int x, int y) {
    fa[find(x)] = find(y);
}

void solve() {
    int n, m, k;
    cin >> n >> m >> k;
    for (int i = 1; i <= n; i++)fa[i] = i;
    for (int i = 1; i <= m; i++) {
        int x, y, w;
        cin >> x >> y >> w;
        node[i] = Node(x, y, w);
    }
    map<int, int> mp1, mp2;
    for (int i = 1; i <= k; i++) {
        int x;
        cin >> x;
        mp1[x]++;
    }

    for (int i = 1; i <= k; i++) {
        int x;
        cin >> x;
        mp2[x]++;
    }

    auto cmp = [&](Node a, Node b) {
        return a.w < b.w;
    };
    sort(node + 1, node + m + 1, cmp);

    int ans = 0;
    int num = 0;
    int A = 0, B = 0;
    for (int i = 1; i <= m; i++) {
        int x = find(node[i].u), y = find(node[i].v);
        if (x == y)continue;
        int w = node[i].w;
        mp1[y] += mp1[x];
        mp2[y] += mp2[x];
        int cnt = min(mp1[y], mp2[y]);
        ans += w * cnt;
        mp1[y] -= cnt;
        mp2[y] -= cnt;
        merge(x, y);
    }
    cout << ans << endl;
}

signed main() {
#ifdef ONLINE_JUDGE
#else
    freopen("test.in", "r", stdin);
    freopen("test.out", "w", stdout);
#endif
    ios::sync_with_stdio(0);
    cout.tie(0);
    cin.tie(0);
    int t = 1;
//    cin >> t;
    while (t--)
        solve();
    return 0;
}