A. Strictly Increasing?

模拟

代码实现 
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
 
using namespace std;
 
int main() {
    int n;
    cin >> n;
    
    vector<int> a(n);
    rep(i, n) cin >> a[i];
    
    rep(i, n-1) {
        if (a[i] >= a[i+1]) {
            puts("No");
            return 0;
        }
    }
    
    puts("Yes");
    
    return 0;
}

B. Make Target

模拟

代码实现 
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
 
using namespace std;
 
int main() {
    int n;
    cin >> n;
    
    vector<string> s(n, string(n, '.'));
    rep(l, n) {
        int r = n-1-l;
        if (l > r) continue;
        char c = '#';
        if (l%2) c = '.';
        for (int i = l; i <= r; ++i) {
            for (int j = l; j <= r; ++j) {
                s[i][j] = c;
            }
        }
    }
    
    rep(i, n) cout << s[i] << '\n';
    
    return 0;
}

C. Shortest Duplicate Subarray

遍历同一个数的相邻两个位置差+1求最小值

代码实现 
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
 
using namespace std;
 
int main() {
    int n;
    cin >> n;
    
    vector<int> a(n);
    rep(i, n) cin >> a[i];
    
    vector<int> cnt(1000005);
    int mult = 0;
    
    const int INF = 1001001001;
    int ans = INF;
    int r = 0;
    rep(l, n) {
        while (r < n and mult == 0) {
            cnt[a[r]]++;
            if (cnt[a[r]] == 2) mult++;
            r++;
        }
        if (mult == 0) break;
        ans = min(ans, r-l);
        if (cnt[a[l]] == 2) mult--;
        cnt[a[l]]--;
    }
    
    if (ans == INF) ans = -1;
    cout << ans << '\n';
    
    return 0;
}

D. Pigeon Swap

考虑将巢里的鸽子看成一个整体，也就是将巢里的鸽子视为放入一个袋子 B 里，然后维护以下三种映射：

  • P→B
  • B→H
  • H→B
代码实现 
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
 
using namespace std;
 
int main() {
    int n, q;
    cin >> n >> q;
    
    vector<int> p2b(n);
    vector<int> b2h(n);
    vector<int> h2b(n);
    rep(i, n) p2b[i] = i;
    rep(i, n) b2h[i] = i;
    rep(i, n) h2b[i] = i;
    
    rep(qi, q) {
        int type;
        cin >> type;
        if (type == 1) {
            int a, b;
            cin >> a >> b;
            --a; --b;
            p2b[a] = h2b[b];
        }
        else if (type == 2) {
            int a, b;
            cin >> a >> b;
            --a; --b;
            swap(h2b[a], h2b[b]);
            b2h[h2b[a]] = a;
            b2h[h2b[b]] = b;
        }
        else {
            int a;
            cin >> a;
            --a;
            int ans = b2h[p2b[a]];
            cout << ans+1 << '\n';
        }
    }
    
    return 0;
}

E. Flip Edge

扩展Dijkstra
第二维用来维护是否翻转
那么只需要开2倍点即可

代码实现 
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
 
using namespace std;
using ll = long long;
using P = pair<ll, int>;
 
int main() {
    int n, m, x;
    cin >> n >> m >> x;
    
    n *= 2;
    vector<vector<int>> to(n);
    rep(i, m) {
        int a, b;
        cin >> a >> b;
        --a; --b;
        to[a*2].push_back(b*2);
        to[b*2+1].push_back(a*2+1);
    }
    
    const ll INF = 1e18;
    priority_queue<P, vector<P>, greater<P>> q;
    vector<ll> dist(n, INF);
    
    auto push = [&](int v, ll d) {
        if (dist[v] <= d) return;
        dist[v] = d;
        q.emplace(d, v);
    };
    push(0, 0);
    while (q.size()) {
        auto [d, v] = q.top(); q.pop();
        if (dist[v] != d) continue;
        for (int u : to[v]) push(u, d+1);
        push(v^1, d+x);
    }
    
    ll ans = min(dist[n-1], dist[n-2]);
    cout << ans << '\n';
    
    return 0;
}

F. Smooth Occlusion

注意到 H 满足单调性（H 可行，H−1 一定也可行）
费用为 ∑i(Ui+Di)−H×N，可以看出 H 越大，费用越小
我们可以二分出满足条件的最大的H

代码实现 
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
 
using namespace std;
using ll = long long;
 
int main() {
    int n, x;
    cin >> n >> x;
    
    vector<int> u(n), d(n);
    rep(i, n) cin >> u[i] >> d[i];
    
    auto judge = [&](ll h) {
        ll l = 0, r = h;
        rep(i, n) {
            ll nl = h-d[i], nr = u[i];
            nl = max(nl, 0ll);
            nr = min(nr, h);
            
            nl = max(nl, l-x);
            nr = min(nr, r+x);
            l = nl; r = nr;
            if (l > r) return false;
        }
        return true;
    };
    
    ll ac = 0, wa = 3e9;
    while (ac+1 < wa) {
        ll wj = (ac+wa)/2;
        if (judge(wj)) ac = wj; else wa = wj;
    }
    
    ll ans = 0;
    rep(i, n) ans += u[i]+d[i];
    ans -= ac*n;
    cout << ans << '\n';
    
    return 0;
}

G. Minimum Steiner Tree 2

steiner树dp+flyod最短路

考虑 a 和 b 这两点间的最短路
记 dp[v][S] 表示在 a→v 的最短路及其路径上挂着的支路中选择若干个点使得 a, S, v 连通的最小费用

时间复杂度为 O(3K⋅N2)

代码实现 
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
 
using namespace std;
using ll = long long;
 
inline void chmin(ll& x, ll y) { if (x > y) x = y; }
 
int main() {
    cin.tie(nullptr) -> sync_with_stdio(false);
    
    int n, k;
    cin >> n >> k;
    
    int k2 = 1<<k;
    vector c(n, vector<ll>(n));
    rep(i, n)rep(j, n) cin >> c[i][j];
    rep(k, n)rep(i, n)rep(j, n) chmin(c[i][j], c[i][k]+c[k][j]);
    
    const ll INF = 1e18;
    vector dp1(n, vector<ll>(k2, INF)); // dp1[v][S]: 以 v 为根且包含点集 S 的最小费用
    vector dp(n, vector(n, vector<ll>(k2, INF))); 
    
    rep(i, n) dp1[i][0] = 0;
    rep(i, k) dp1[i][1<<i] = 0;
    rep(s, k2) {
        for (int t = s; t; t = (t-1)&s) {
            rep(i, n) chmin(dp1[i][s], dp1[i][t]+dp1[i][s^t]);
        }
        rep(i, n)rep(j, n) chmin(dp1[i][s], dp1[j][s]+c[j][i]);
    }
    
    rep(sv, n) {
        auto& ndp = dp[sv];
        ndp[sv][0] = 0;
        rep(s, k2) {
            for (int t = s; t; t = (t-1)&s) {
                rep(i, n) chmin(ndp[i][s], dp1[i][t]+ndp[i][s^t]);
            }
            rep(i, n)rep(j, n) chmin(ndp[i][s], ndp[j][s]+c[j][i]);
        }
    }
    
    int q;
    cin >> q;
    rep(qi, q) {
        int a, b;
        cin >> a >> b;
        --a; --b;
        ll ans = dp[a][b][k2-1];
        cout << ans << '\n';
    }
    
    return 0;
}