AtCoder Beginner Contest 345

A - Leftrightarrow

#include <bits/stdc++.h>

using namespace std;

using i32 = int32_t;
using i64 = long long;
using ldb = long double;

#define int long long

using vi = vector<int>;
using pii = pair<int, int>;

const int mod = 1e9 + 7;
const int inf = 1e18;

i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    string s;
    cin >> s;
    if (s.front() == '<' and s.back() == '>' and s.substr(1, s.size() - 2) == string(s.size() - 2, '=')) {
        cout << "Yes\n";
    } else {
        cout << "No\n";
    }
    return 0;
}

B - Integer Division Returns

#include <bits/stdc++.h>

using namespace std;

using i32 = int32_t;
using i64 = long long;
using ldb = long double;

#define int long long

using vi = vector<int>;
using pii = pair<int, int>;

const int mod = 1e9 + 7;
const int inf = 1e18;

i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int x;
    cin >> x;
    if(x > 0 ) cout << ( x + 9 ) / 10 << "\n";
    else cout << x / 10 << "\n";
    return 0;
}

C - One Time Swap

#include <bits/stdc++.h>

using namespace std;

using i32 = int32_t;
using i64 = long long;
using ldb = long double;

#define int long long

using vi = vector<int>;
using pii = pair<int, int>;

const int mod = 1e9 + 7;
const int inf = 1e18;

i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    map<char, int> cnt;
    string s;
    cin >> s;
    for (auto i: s) cnt[i]++;
    int sum = s.size(), res = 0, f = 0;
    for (auto i: s) {
        sum--, cnt[i]--;
        res += sum - cnt[i], f |= cnt[i] > 0;
    }
    cout << res + f << "\n";
    return 0;
}

D - Tiling

可以直接暴搜，但是暴搜也是要有一些技巧

#include <bits/stdc++.h>

using namespace std;

using i32 = int32_t;
using i64 = long long;
using ldb = long double;

using vi = vector<int>;
using pii = pair<int, int>;

const int mod = 998244353;
const int inf = 1e18;

const vi dx = {0, 0, 1, -1}, dy = {1, -1, 0, 0};

i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n, h, w;
    cin >> n >> h >> w;
    vi a(n), b(n);
    for (int i = 0; i < n; i++) cin >> a[i] >> b[i];
    vector f(h, vi(w));
    vi vis(n);
    auto dfs = [&](auto self, int x, int y) -> void {
        if (x == h) {
            cout << "Yes\n";
            exit(0);
        }
        if (y == w) return self(self, x + 1, 0);
        if (f[x][y]) return self(self, x, y + 1);
        for (int i = 0; i < n; i++) {
            if (vis[i]) continue;
            vis[i] = 1;
            for (int t = 0, ok; t < 2; t++) {
                swap(a[i], b[i]);
                if( x + a[i] > h or y + b[i] > w ) continue;
                ok = 1;
                for (int u = 0; ok and u < a[i]; u++)
                    for (int v = 0; ok and v < b[i]; v++)
                        if (f[x + u][y + v]) ok = 0;
                if (not ok) continue;
                for (int u = 0; u < a[i]; u++)
                    for (int v = 0; v < b[i]; v++)
                        f[x + u][y + v] = 1;
                self(self, x, y + 1);
                for (int u = 0; u < a[i]; u++)
                    for (int v = 0; v < b[i]; v++)
                        f[x + u][y + v] = 0;
            }
            vis[i] = 0;
        }
    };
    dfs(dfs, 0, 0);
    cout << "No\n";
    return 0;
}

E - Colorful Subsequence

有 \(N\) 个球，每一个球有一个颜色 \(C_i\) 和价值 \(V_i\)，现在要删除其中的 \(K\) 个球，使得剩下的球没有相邻的两个球颜色相等。求剩下的球的最大价值总和。

设\(f[i][j]\)表示为前\(i\)个球删掉\(j\)个且钦定保留第\(i\)个球的最大价值总和。

假设剩下\(s\)个球，则$s = i - j \(，显然可以得到\)s\le i \le s + k\(，如果我们枚举出了\)i,s$就可以得到转移

\[f[i][i-s] = \max_{s \le p \le i, C_p\ne C_i}(f[p-1][p-s]) + v[i] \]

比如我们枚举出了\(i=4,s=1\)，则我们要求的就是\(f[4][3]=\max(f[3][3] , f[2][2], f[1][1],f[0][0]) + v[4]\)

所以其实我们可以统计出前缀最大值，以及与前缀最大值不同的前缀次大值的，然后根据颜色是否相同判断选择那个即可。

#include <bits/stdc++.h>

using namespace std;

using i32 = int32_t;
using i64 = long long;
using ldb = long double;

#define int long long

using vi = vector<int>;
using pii = pair<int, int>;

const int mod = 998244353;
const int inf = 1e18;

const vi dx = {0, 0, 1, -1}, dy = {1, -1, 0, 0};

i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n, k;
    cin >> n >> k;
    vi C(n + 2), V(n + 2);
    for (int i = 1; i <= n; i++) cin >> C[i] >> V[i];
    C[0] = n + 1, C[n + 1] = n + 2;
    n += 2;
    vector f(n, vi(k + 1, -inf));
    f[0][0] = 0;
    for (int s = 1; s <= n - 1 - k; s++) {
        pii mx[2] = {{-inf, 0},// 最大值
                     {-inf, 0}}; // 与最大值颜色不同的 次大值
        for (int i = s; i <= s + k; i++) {
            pii v = {f[i - 1][i - s], C[i - 1]};
            if (v > mx[0])
                swap(v, mx[0]);
            if (mx[0].second == mx[1].second || (v > mx[1] and v.second != mx[0].second))
                mx[1] = v;
            f[i][i - s] = mx[mx[0].second == C[i]].first + V[i];
        }
    }
    cout << max(-1ll, f[n - 1][k]) << "\n";
    return 0;
}