Codeforces Round #731 (Div. 3)

Codeforces Round #731 (Div. 3)

A - Shortest Path with Obstacle

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
 
    int _;
    for (cin >> _; _; --_) {
        pair<int, int> pos[3];
        for (int i = 0; i < 3; ++i)
            cin >> pos[i].first >> pos[i].second;
 
        if (pos[0].first == pos[1].first && pos[0].first == pos[2].first) {
            if (pos[2].second > pos[0].second && pos[2].second < pos[1].second)
                cout << pos[1].second - pos[0].second + 2 << '
';
            else if (pos[2].second > pos[1].second && pos[2].second < pos[0].second)
                cout << pos[0].second - pos[1].second + 2 << '
';
            else
                cout << abs(pos[0].second - pos[1].second) << '
';
        } else if (pos[0].second == pos[1].second && pos[0].second == pos[2].second) {
            if (pos[2].first > pos[0].first && pos[2].first < pos[1].first)
                cout << pos[1].first - pos[0].first + 2 << '
';
            else if (pos[2].first > pos[1].first && pos[2].first < pos[0].first)
                cout << pos[0].first - pos[1].first + 2 << '
';
            else
                cout << abs(pos[0].first - pos[1].first) << '
';
        }
        else cout << abs(pos[0].first - pos[1].first) + abs(pos[0].second - pos[1].second) << '
';
    }
    return 0;
}

B - Alphabetical Strings

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
 
    int _;
    for (cin >> _; _; --_) {
        string s;
        cin >> s;
 
        deque<char> du;
        for (auto &c : s)
            du.emplace_back(c);
 
        bool f = 1;
        for (int i = s.size() - 1; !du.empty() && f; --i)
            if (du.front() == 'a' + i)
                du.pop_front();
            else if (du.back() == 'a' + i)
                du.pop_back();
            else
                f = 0;
 
        cout << (f ? "YES" : "NO") << '
';
    }
    return 0;
}

C - Pair Programming

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
 
    int _;
    for (cin >> _; _; --_) {
        int n, m, k;
        cin >> k >> n >> m;
 
        queue<int> x, y, ans;
        for (int i = 0; i < n; ++i) {
            int a;
            cin >> a;
            x.emplace(a);
        }
        for (int i = 0; i < m; ++i) {
            int a;
            cin >> a;
            y.emplace(a);
        }
 
        while (!x.empty() || !y.empty()) {
            if (!x.empty() && x.front() == 0) {
                ans.emplace(0);
                x.pop();
                ++k;
            } else if (!y.empty() && y.front() == 0) {
                ans.emplace(0);
                y.pop();
                ++k;
            } else if (!x.empty() && x.front() <= k) {
                ans.emplace(x.front());
                x.pop();
            } else if (!y.empty() && y.front() <= k) {
                ans.emplace(y.front());
                y.pop();
            } else
                break;
        }
 
        if (!x.empty() || !y.empty())
            cout << "-1
";
        else {
            while (!ans.empty()) {
                cout << ans.front() << ' ';
                ans.pop();
            }
            cout << '
';
        }
    }
    return 0;
}

D - Co-growing Sequence

(b_0 = 1)， 剩下的模拟

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
 
    int _;
    for (cin >> _; _; --_) {
        int n;
        cin >> n;
 
        vector<int> a(n);
        for (auto &i : a)
            cin >> i;
 
        for (int i = 0, ls = a[0]; i < n; ++i) {
            int cur = 0;
            for (int j = 29; ~j; --j)
                if ((ls >> j & 1) && !(a[i] >> j & 1))
                    cur ^= 1 << j;
            cout << cur << ' ';
            ls = a[i] ^ cur;
        }
        cout << '
';
    }
    return 0;
}

E - Air Conditioners

考虑在空调右边的各自的温度为

(i - pos[j] + tem[j] = i + (tem[j] - pos[j]))

扫一遍维护(min(tem[j] - pos[j]))即可

空调在左边同理维护空调位置和空调温度的和即可， 最后取个最小值

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
 
    int _;
    for (cin >> _; _; --_) {
        int n, k;
        cin >> n >> k;
 
        vector<pair<int, int>> b(k);
        for (auto &i : b)
            cin >> i.first;
        for (auto &i : b)
            cin >> i.second;
        sort(b.begin(), b.end());
 
        vector<int> ans(n + 1, 2e9);
        for (int i = 1, j = 0, g = -1; i <= n; ++i)
            if (~g) {
                if (j < k && b[j].first == i) {
                    if (b[j].second - i < b[g].second - b[g].first)
                        g = j;
                    ++j;
                }
                ans[i] = i + b[g].second - b[g].first;
            } else if (j < k && b[j].first == i) {
                g = j++;
                ans[i] = i + b[g].second - b[g].first;
            }
 
        for (int i = n, j = k - 1, g = -1; i; --i) 
            if (~g) {
                if (j != -1 && b[j].first == i) {
                    if (i + b[j].second < b[g].second + b[g].first)
                        g = j;
                    --j;
                }
                ans[i] = min(ans[i], b[g].second + b[g].first - i);
            } else if (j != -1 && b[j].first == i) {
                g = j--;
                ans[i] = min(ans[i], b[g].second + b[g].first - i);
            }
 
        for (int i = 1; i <= n; ++i)
            cout << ans[i] << char(" 
"[i == n]);
    }
    return 0;
}

F - Array Stabilization (GCD version)

明显的二分题， 主要是怎么区间求gcd，用线段树即可

记得把数组复制一倍，防止越界

struct BIT {
    int val[N << 2];
    void push_up(int p) {
        val[p] = __gcd(val[p << 1], val[p << 1 | 1]);
    }
    void build(int p, int l, int r, vector<int> &a) {
        if (l == r) {
            val[p] = a[l] - a[l - 1];
            return;
        }
        int mid = l + r >> 1;
        build(p << 1, l, mid, a);
        build(p << 1 | 1, mid + 1, r, a);
        push_up(p);
    }
    int ask(int p, int l, int r, int L, int R) {
        if (L <= l && r <= R)
            return val[p];
        int mid = l + r >> 1;
        if (mid >= R)
            return ask(p << 1, l, mid, L, R);
        if (mid < L)
            return ask(p << 1 | 1, mid + 1, r, L, R);
        return __gcd(ask(p << 1, l, mid, L, R), ask(p << 1 | 1, mid + 1, r, L, R));
    }
} bit;
 
bool check(int mid, vector<int>& a) {
    if (mid == 0) {
        for (int n = a.size() >> 1, i = 2; i <= n; ++i)
            if (a[i] ^ a[1])
                return 0;
        return 1;
    }
 
    int n = a.size() - 1, ls = abs(__gcd(a[1], bit.ask(1, 1, n, 2, mid + 1)));
    for (int i = 2; i <= n >> 1; ++i) {
        int cur = abs(__gcd(a[i], bit.ask(1, 1, n, i + 1, i + mid)));
        if (cur ^ ls)
            return 0;
    }
    return 1;
}
 
int main() {
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
 
    int _;
    for (cin >> _; _; --_) {
        int n;
        cin >> n;
 
        vector<int> a((n << 1) + 1);
        for (int i = 1; i <= n; ++i) {
            cin >> a[i];
            a[i + n] = a[i];
        }
 
        bit.build(1, 1, n << 1, a);
 
        int l = 0, r = n - 1;
        while (l < r) {
            int mid = l + r >> 1;
            if (check(mid, a))
                r = mid;
            else
                l = mid + 1;
        }
        cout << l << '
';
    }
    return 0;
}

G - How Many Paths?

有向图缩点建新图， 跑个拓扑图即可

int _, n, m, ans[N];
int c[N], scnt, cnt[N];
int dfn[N], low[N], df, st[N], top;
int deg[N];
bool inst[N], tag[N];
vector<int> h[N], nh[N];
 
void tarjan(int x) {
    dfn[x] = low[x] = ++df;
    inst[st[++top] = x] = 1;
 
    for (auto &y : h[x])
        if (!dfn[y]) {
            tarjan(y);
            low[x] = min(low[x], low[y]);
        }
        else if (inst[y])
            low[x] = min(low[x], dfn[y]);
 
    if (dfn[x] == low[x]) {
        ++scnt;
        int z;
        do {
            inst[z = st[top--]] = 0;
            c[z] = scnt;
            ++cnt[scnt];
        } while (z != x);
    }
}
 
int main() {
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
 
    for (cin >> _; _; --_) {
        cin >> n >> m;
 
        for (int i = 1; i <= n; ++i) {
            h[i].clear();
            ans[i] = cnt[i] = tag[i] = c[i] = dfn[i] = 0;
        }
 
        for (int i = 1; i <= m; ++i) {
            int u, v;
            cin >> u >> v;
            if (u ^ v)
                h[u].emplace_back(v);
            else
                tag[u] = 1;
        }
 
        top = scnt = df = 0;
        tarjan(1);
 
        for (int i = 1; i <= scnt; ++i)
            nh[i].clear();
        for (int i = 1; i <= n; ++i)
            if (c[i])
                for (auto &y : h[i])
                    if (c[y] && c[i] != c[y]) {
                        nh[c[i]].emplace_back(c[y]);
                        ++deg[c[y]];
                    }
 
        ans[c[1]] = 1;
        for (int i = 1; i <= n; ++i)
            if (c[i] && tag[i])
                ans[c[i]] = -1;
 
        queue<int> q;
        q.push(c[1]);
        while (!q.empty()) {
            int x = q.front();
            q.pop();
 
            if (cnt[x] > 1)
                ans[x] = -1;
 
            for (auto &y : nh[x]) {
                if (--deg[y] == 0)
                    q.push(y);
 
                if (ans[x] == -1)
                    ans[y] = -1;
                else if (ans[y] != -1) {
                    ans[y] = min(2, ans[y] + 1);
                    if (ans[x] == 2)
                        ans[y] = 2;
                }
            }
        }
 
        for (int i = 1; i <= n; ++i)
            cout << ans[c[i]] << char(" 
"[i == n]);
    }
    return 0;
}