ABC338

T1：Capitalized?

模拟

代码实现

s = input()
if s.istitle():
    print('Yes')
else:
    print('No')

T2：Frequency

模拟

代码实现

#include <bits/stdc++.h>
 
using namespace std;
 
int main() {
    string s;
    cin >> s;
    
    vector<int> cnt(256);
    for (char c : s) cnt[c]++;
    
    pair<int, char> ans(0, 'a');
    for (char c = 'a'; c <= 'z'; ++c) {
        ans = min(ans, make_pair(-cnt[c], c));
    }
    
    cout << ans.second << '\n';
    
    return 0;
}

T3：Leftover Recipes

枚举做多少份料理 $A$，然后算出剩下的材料最多可以做多少份料理 $B$ 即可

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> q(n), a(n), b(n);
    rep(i, n) cin >> q[i];
    rep(i, n) cin >> a[i];
    rep(i, n) cin >> b[i];
    
    int ans = 0;
    for (int x = 0;; ++x) {
        vector<int> r(n);
        rep(i, n) r[i] = q[i] - a[i]*x;
        bool ok = true;
        rep(i, n) if (r[i] < 0) ok = false;
        if (!ok) break;
        int y = 1001001001;
        rep(i, n) {
            if (b[i] == 0) continue;
            y = min(y, r[i]/b[i]);
        }
        ans = max(ans, x+y);
    }
    
    cout << ans << '\n';
    
    return 0;
}

T4：Island Tour

考虑“如果断开哪座桥，从 $X_i$ 到 $X_{i+1}$ 分别需要走的步数” ，注意到断开从 $X_i$ 到 $X_{i+1}$ 的最短路上的任意一座桥，其所需走的步数是一样的，类似地，对于断开剩下的桥中任意一个所需走的步数也是一样的， 所以用差分来算它们的前缀和即可！

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;
 
int main() {
    int n, m;
    cin >> n >> m;
    
    vector<int> x(m);
    rep(i, m) {
        cin >> x[i];
        x[i]--;
    }
    
    vector<ll> d(n+1);
    auto add = [&](int l, int r, int x) {
        d[l] += x;
        d[r] -= x;
    };
    rep(i, m-1) {
        int s = x[i], t = x[i+1];
        if (s > t) swap(s, t);
        int a = t-s, b = n-a;
        add(0, s, a);
        add(s, t, b);
        add(t, n, a);
    }
    
    rep(i, n) d[i+1] += d[i];
    d.pop_back();
    
    ll ans = ranges::min(d);
    cout << ans << '\n';
    
    return 0;
}

T5：Chords

本质上是括号匹配问题，用栈来维护即可

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;
 
int main() {
    int n;
    cin >> n;
    
    vector<int> d(n*2);
    rep(i, n) {
        int a, b;
        cin >> a >> b;
        --a; --b;
        d[a] = i; d[b] = i;
    }
    
    stack<int> st;
    for (int x : d) {
        if (st.size() and st.top() == x) {
            st.pop();
        }
        else st.push(x);
    }
    
    if (st.size() == 0) puts("No");
    else puts("Yes");
    
    return 0;
}

T6：Negative Traveling Salesman

floyd + 状压dp
记 dp[S][i] 表示已经遍历过的点集为 $S$ 且当前在点 $i$ 处的最小费用

代码实现

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
 
using namespace std;
 
inline void chmin(int& a, int b) { if (a > b) a = b; }
 
int main() {
    int n, m;
    cin >> n >> m;
    
    const int INF = 1001001001;
    vector g(n, vector<int>(n, INF));
    rep(i, n) g[i][i] = 0;
    rep(i, m) {
        int a, b, c;
        cin >> a >> b >> c;
        --a; --b;
        g[a][b] = min(g[a][b], c);
    }
    
    rep(k, n)rep(i, n)rep(j, n) chmin(g[i][j], g[i][k]+g[k][j]);
    
    int n2 = 1<<n;
    vector dp(n2, vector<int>(n, INF));
    rep(i, n) dp[1<<i][i] = 0;
    rep(s, n2)rep(i, n) if (dp[s][i] != INF) {
        rep(j, n) if ((~s>>j&1) and g[i][j] != INF) {
            int ns = s|1<<j;
            chmin(dp[ns][j], dp[s][i]+g[i][j]);
        }
    }
    
    int ans = ranges::min(dp[n2-1]);
    if (ans == INF) puts("No");
    else cout << ans << '\n';
    
    return 0;
}