AtCoder Beginner Contest 327

1|0A - ab

#include<bits/stdc++.h>

using namespace std;


#define mp make_pair
#define int long long
using vi = vector<int>;
using pii = pair<int, int>;

const int inf = 1e9;

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n;
    string s;
    cin >> n >> s;
    for( int i = 1 ; i < n ; i ++ ){
        if( s[i] == 'a' and s[i-1] == 'b' ) {
            cout << "Yes\n";
            return 0;
        }else if( s[i] == 'b' and s[i-1] == 'a' ){
            cout << "Yes\n";
            return 0;
        }
    }
    cout << "No\n";
    return 0;
}

2|0B - A^A

b = int(input())
a = 1

while (a ** a) < b:
    a += 1

if (a ** a) == b:
    print(a)
else:
    print(-1)

3|0C - Number Place

#include<bits/stdc++.h>

using namespace std;

#define mp make_pair
#define int long long
using vi = vector<int>;
using pii = pair<int, int>;

const int inf = 1e9;

int power(int x, int y) {
    int ans = 1;
    while (y) {
        if (y & 1) ans = ans * x;
        x = x * x, y /= 2;
    }
    return ans;
}

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n = 9;
    vector a(n, vi(n));

    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            cin >> a[i][j];

    for (int i = 0; i < n; i++) {
        set<int> vis;
        for (int j = 0; j < n; j++)
            vis.insert(a[i][j]);
        if (vis.size() != n) {
            cout << "No\n";
            return 0;
        }
    }
    for (int i = 0; i < n; i++) {
        set<int> vis;
        for (int j = 0; j < n; j++)
            vis.insert(a[j][i]);
        if (vis.size() != n) {
            cout << "No\n";
            return 0;
        }
    }
    for (int i = 0; i < n; i += 3) {
        for (int j = 0; j < n; j += 3) {
            set<int> vis;
            for (int x = 0; x < 3; x++)
                for (int y = 0; y < 3; y++)
                    vis.insert(a[i + x][j + y]);
            if (vis.size() != n) {
                cout << "No\n";
                return 0;
            }
        }
    }
    cout << "Yes\n";
    return 0;
}

4|0D - Good Tuple Problem

赛时的思考还是有些欠缺，一看到这道题，就直接抄了个 2-SAT 的板子，结果还 wa 了一发。

#include<bits/stdc++.h>

using namespace std;


#define mp make_pair
#define int long long
using vi = vector<int>;
using pii = pair<int, int>;

const int inf = 1e9;

int N;
vector<vi> e;
vi dfn, inStk, low, scc;
int cnt, sc;
stack<int> stk;

void tarjan(int x) {
    low[x] = dfn[x] = ++cnt;
    inStk[x] = 1, stk.push(x);
    for (auto y: e[x]) {
        if (!dfn[y]) {
            tarjan(y), low[x] = min(low[x], low[y]);
        } else if (inStk[y]) {
            low[x] = min(low[x], dfn[y]);
        }
    }
    if (low[x] == dfn[x]) {
        sc++;
        for (int y; true;) {
            y = stk.top(), stk.pop();
            inStk[y] = 0, scc[y] = sc;
            if (x == y) break;
        }
    }
    return;
}

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n, m;
    cin >> n >> m;
    vector<int> a(m + 1), b(m + 1);
    for (int i = 1; i <= m; i++) cin >> a[i];
    for (int i = 1; i <= m; i++) cin >> b[i];

    N = n * 2 + 2;
    e = vector<vi>(N);
    dfn = inStk = low = scc = vi(N);

    for (int i = 1; i <= m; i++) {
        if (a[i] == b[i]) {
            cout << "No\n";
            return 0;
        }
        e[2 * a[i]].push_back(2 * b[i] + 1);
        e[2 * a[i] + 1].push_back(2 * b[i]);
    }
    for (int i = 2; i <= n * 2 + 1; i++)
        if (!dfn[i])
            tarjan(i);

    for (int i = 1; i <= n; i++) {
        if (scc[i * 2] == scc[i * 2 + 1]) {
            cout << "No\n";
            return 0;
        }
    }
    cout << "Yes\n";
    return 0;
}

其实认真思考后，发现符合条件的情况一定是二分图的，所以直接进行黑白染色就好了。

#include<bits/stdc++.h>

using namespace std;

#define mp make_pair
#define int long long
using vi = vector<int>;
using pii = pair<int, int>;
using ldb = long double;
const int inf = 1e9;

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);

    int n, m;
    cin >> n >> m;
    vector<vi> e(n);
    vector<int> a(m), b(m);
    for (auto &i: a) cin >> i, i--;
    for (auto &i: b) cin >> i, i--;
    for (int i = 0; i < m; i++)
        e[a[i]].push_back(b[i]), e[b[i]].push_back(a[i]);
    vi v(n + n, -1);
    auto dfs = [&v, e](auto &&self, int x, int w) -> void {
        for (auto y: e[x]) {
            if (v[y] == -1) {
                v[y] = w ^ 1, self(self, y, v[y]);
            } else if (v[y] == w) {
                cout << "No\n";
                exit(0);
            }
        }
    };

    for (int i = 0; i < n; i++) {
        if (v[i] >= 0) continue;
        v[i] = 0;
        dfs(dfs, i, 0);
    }
    cout << "Yes\n";
    return 0;
}

当然也可以用判定二分图的方式。

5|0E - Maximize Rating

观察式子，发现当$k$确定时，只有$\sum (0.9)^{k-i} Q_i$是变换的，其他部分都是常量。对于这个值，可以直接背包求一下。

#include<bits/stdc++.h>

using namespace std;


#define mp make_pair
#define int long long
using vi = vector<int>;
using pii = pair<int, int>;
using ldb = long double;
const int inf = 1e9;

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n;
    cin >> n;
    vector<ldb> a(n + 1), q(n + 1), s(n + 1), t(n + 1);
    for (int i = n; i >= 1; i--) cin >> a[i];
    q[1] = 1;
    for (int i = 2; i <= n; i++) q[i] = q[i - 1] * 0.9;
    for (int i = 1; i <= n; i++) s[i] = s[i - 1] + q[i];
    for (int i = 1; i <= n; i++) t[i] = 1200.0 / sqrt((ldb) i);

    vector<vector<ldb>> f(n + 1, vector<ldb>(n + 1, -1E18));
    ldb res = -1E18;
    f[0][0] = 0;
    for (int i = 1; i <= n; i++) {
        f[i][0] = 0;
        for (int j = 1; j <= i; j++) {
            f[i][j] = max(f[i - 1][j], f[i - 1][j - 1] + q[j] * a[i]);
            res = max(res, f[i][j] / s[j] - t[j]);
        }
    }
    cout << fixed << setprecision(10) << res << "\n";

    return 0;
}

6|0F - Apples

其实本质上是个二维数点问题，问用$w\times d$的矩形最多可以框选多少个点。

用一种类似扫描线的方式在$t$时给$x$加一，然后在$t+d$时给$x$减一，这样可以优化掉一维，对于剩下的一维可以采用类似双指针的方式此时复杂度是$O(N^2)$的依旧不行。

我们考虑每次扫描线后每次都是考虑宽度$w$范围有多少个点，我们不如反过来思考，考虑每个点对那些区间产生贡献，答案是$[x,x+w)$，所以每次对这个区间加一，询问时只要询问全局最优解就好了。

#include<bits/stdc++.h>

using namespace std;

using pii = pair<int, int>;

struct Node {
    int l, r, maxVal, lazy;
    Node *left, *right;

    Node(int l, int r, Node *left, Node *right)
            : l(l), r(r), left(left), right(right) {
        maxVal = lazy = 0;
    }
} *root;

Node *build(int l, int r) {
    if (l == r) return new Node(l, r, nullptr, nullptr);
    int mid = (l + r) / 2;
    auto left = build(l, mid), right = build(mid + 1, r);
    return new Node(l, r, left, right);
}

void pushDown(Node *rt) {
    rt->left->maxVal += rt->lazy, rt->left->lazy += rt->lazy;
    rt->right->maxVal += rt->lazy, rt->right->lazy += rt->lazy;
    rt->lazy = 0;
    return;
}

void modify(Node *rt, int l, int r, int v) {
    if (l > rt->r or r < rt->l) return;
    if (l <= rt->l and rt->r <= r) {
        rt->maxVal += v, rt->lazy += v;
        return;
    }
    pushDown(rt);
    int mid = (rt->l + rt->r) / 2;
    if (l <= mid) modify(rt->left, l, r, v);
    if (r > mid) modify(rt->right, l, r, v);
    rt->maxVal = max(rt->left->maxVal, rt->right->maxVal);
    return;
}

const int N = 4e5;


int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n, d, w;
    cin >> n >> d >> w;

    vector<vector<pii> > f(N + 1);
    for (int t, x, i = 1; i <= n; i++) {
        cin >> t >> x;
        f[t].emplace_back(x, 1);
        if (t + d <= N) f[t + d].emplace_back(x, -1);
    }
    root = build(1, N);
    int res = 0;
    for (int x = 1; x <= N; x++) {
        for (const auto &[y, v]: f[x])
            modify(root, y, min(N, y + w - 1), v);
        res = max(res, root->maxVal);
    }
    cout << res << "\n";
    return 0;
}

__EOF__

本文作者：PHarr
本文链接：https://www.cnblogs.com/PHarr/p/17817706.html
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