2024牛客寒假算法基础集训营1

2024牛客寒假算法基础集训营1

A.DFS搜索

思路：看dfs其实就是一道字符题目

#include <bits/stdc++.h>

using namespace std;
string x = "dfs";
string y = "DFS";

void solve() {
    int n;
    cin >> n;
    string st;
    cin >> st;
    int xx = 0, yy = 0;
    for (int i = 0; i < n; i++) {
        if (st[i] == x[xx]) xx++;
        if (st[i] == y[yy]) yy++;
    }
    if (xx == 3) xx = 1;
    else xx = 0;
    if (yy == 3) yy = 1;
    else yy = 0;
    cout << (yy) << " " << (xx) << endl;
}

int main() {
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
}

B.关鸡

思路：1.最多就3个就是(0,1)(0,-1)(1,0) 2.左右怎么才能不出去? 左右都被堵死 3.如何堵死? 左右同一根线或者相距一格

#include <bits/stdc++.h>

using namespace std;
#define int long long

void solve() {
    int n;
    cin >> n;
    if (n == 0) {
        cout << "3" << endl;
        return;
    }
    set<pair<int, int>> q;
    for (int i = 0; i < n; i++) {
        int a, b;
        cin >> a >> b;
        q.insert({a - 1, b});
    }

    int res = 4;
    int ll = 2;
    int rr = 2;
    int a = q.count({0, -1}), b = q.count({0, 1}), c = q.count({1, 0});
    if (c) {
        ll = 1;
        rr = 1;
    }
    if (a || b || c) {
        res = min(res, 2ll);
        if (a && b || a && c || b && c) {
            res = min(res, 1ll);
        }
        if (a && b && c) {
            res = 0;
        }
    }
    for (auto [r, c]: q) {
        if (c < 0) {
            if (q.count({r ^ 1, c - 1}) || q.count({r ^ 1, c + 1}) || q.count({r ^ 1, c})) {
                ll = 0;
            } else {
                ll = min(ll, 1ll);
            }
        } else if (c > 0) {
            if (q.count({r ^ 1, c - 1}) || q.count({r ^ 1, c + 1}) || q.count({r ^ 1, c})) {
                rr = 0;
            } else {
                rr = min(rr, 1ll);
            }
        }
    }
    res = min(res, ll + rr);
    cout << res << endl;
}

signed main() {
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
}                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                        

C.按闹分配

思路：很明显不满意度最小的时候就是前面时间短，后面时间长，可以用前缀和来维护一下
$$
S_c-S_min=n-i(假设他是第i个去的)
$$

#include <bits/stdc++.h>

using namespace std;
#define int long long

signed main() {
    int n, Q, tc;
    cin >> n >> Q >> tc;
    vector<int> t(n);
    vector<int> s(n + 2);
    for (int i = 0; i < n; i++) {
        cin >> t[i];
    }
    sort(t.begin(), t.end());
    for (int i = 0; i < n; i++) {
        s[i + 1] = s[i] + t[i];
    }
    while (Q--) {
        int m;
        cin >> m;
        int c = min(n, m / tc);
        int ans = s[n - c] + tc;
        cout << ans << endl;
    }
}

E.本题又主要考查了贪心

思路：这题mn很小，根本不用想就狠狠暴力就行(debug搞了好久、、)，后面发现是我后面ans的问题、、、

#include <bits/stdc++.h>

using namespace std;
int m, n, num, ans;
const int MAX = 1e6;
int a[MAX], v[MAX], u[MAX];


void dfs(int x) {
    if (x == num) {
        int res = 0;
        for (int i = 1; i < n; i++) {
            if (a[i] > a[0])res++;
        }
        ans = min(ans, res);
        return;
    }
    a[u[x]] += 3;
    dfs(x + 1);
    a[u[x]] -= 3;
    a[u[x]] += 1;
    a[v[x]] += 1;
    dfs(x + 1);
    a[u[x]] -= 1;
    a[v[x]] -= 1;
    a[v[x]] += 3;
    dfs(x + 1);
    a[v[x]] -= 3;
}

void solve() {
    cin >> n >> m;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    num = 0;
    for (int i = 0; i < m; i++) {
        int u1, v1;
        cin >> u1 >> v1;
        if (u1 == 1 || v1 == 1)a[0] += 3;
        else {
            u[num] = u1 - 1;
            v[num] = v1 - 1;
            num++;
        }
    }
    ans = 1e6;
    dfs(0);
    cout << min(ans + 1, n) << endl;
}

signed main() {
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
}

G.why买外卖

思路：只要超过就能用，那假设都能用，怎么会出现不能用的？ 减去还不行的那种

#include <bits/stdc++.h>

using namespace std;
#define int long long

void solve() {
    int n, m;
    cin >> n >> m;
    vector<pair<int, int>> q(n + 1);
    for (int i = 0; i < n; i++) {
        int a, b;
        cin >> a >> b;
        q[i].first = a;
        q[i].second = b;
        m+=b;
    }
    sort(q.rbegin(), q.rend());
    for(int i=0;i<n;i++){
        if(m>=q[i].first) break;
        else{
            m-=q[i].second;
        }
    }
    cout << m << endl;
}

signed main() {
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
}

L.要有光

思路：纯纯几何题，画个图推一下就行

#include <bits/stdc++.h>

using namespace std;

void solve() {
    int c, d, h, w;
    cin >> c >> d >> h >> w;
    cout << 3 * c * w << endl;
}

int main() {
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
}

M.牛客老粉才知道的秘密

思路：就是一个找规律的题就是看与6的关系

#include <bits/stdc++.h>

using namespace std;
#define int long long

void solve() {
    int n;
    cin >> n;
    if (n % 6 == 0) {
        cout << n / 6 << "\n";
        return;
    }
    int x = n / 6 + 1;
    x = x * 2 - 2;
    cout << x << endl;
}

signed main() {
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
}