Codeforces Round #646 (Div. 2)

传送门

视频题解。

A. Odd Selection

注意到(nleq 1000)，所以直接暴力枚举即可。

Code

/*
 * Author:  heyuhhh
 * Created Time:  2020/5/31 22:35:54
 */
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#include <functional>
#include <numeric>
#define MP make_pair
#define fi first
#define se second
#define pb push_back
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define Local
#ifdef Local
  #define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
  void err() { std::cout << std::endl; }
  template<typename T, typename...Args>
  void err(T a, Args...args) { std::cout << a << ' '; err(args...); }
  template <template<typename...> class T, typename t, typename... A> 
  void err(const T <t> &arg, const A&... args) {
  for (auto &v : arg) std::cout << v << ' '; err(args...); }
#else
  #define dbg(...)
#endif
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 1e5 + 5;
 
void run() {
    int n, x; cin >> n >> x;
    vector <int> a(n);
    int cnt[2] = {0, 0};
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        if (a[i] & 1) ++cnt[1];
        else ++cnt[0];
    }
    for (int i = 1; i <= x; i += 2) {
        if (cnt[1] >= i && cnt[0] >= x - i) {
            cout << "YES" << '
';
            return;
        }
    }
    cout << "NO" << '
';
}
 
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    cout << fixed << setprecision(20);
    int T; cin >> T; while(T--)
    run();
    return 0;
}

B. Subsequence Hate

最后的形式一定为(000...111,111...000,1111..11,00..00)这几种形式，所以只用考虑这几种情况就行。前两种枚举一下即可解决，后两种也很简单。

Code

/*
 * Author:  heyuhhh
 * Created Time:  2020/5/31 22:42:37
 */
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#include <functional>
#include <numeric>
#define MP make_pair
#define fi first
#define se second
#define pb push_back
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define Local
#ifdef Local
  #define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
  void err() { std::cout << std::endl; }
  template<typename T, typename...Args>
  void err(T a, Args...args) { std::cout << a << ' '; err(args...); }
  template <template<typename...> class T, typename t, typename... A> 
  void err(const T <t> &arg, const A&... args) {
  for (auto &v : arg) std::cout << v << ' '; err(args...); }
#else
  #define dbg(...)
#endif
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 1e5 + 5;
 
void run() {
    string s; cin >> s;
    int n = s.length();
    int ans = INF;
    vector <int> sum(n);
    for (int i = 0; i < n; i++) {
        if (i == 0) sum[0] = (s[i] == '1');
        else sum[i] = sum[i - 1] + (s[i] == '1');    
    }
    int cnt0 = 0, cnt1 = 0;
    for (int i = n - 1; i; i--) {
        if (s[i] == '0') {
            ++cnt0;
        } else {
            ++cnt1;
        }
        ans = min(ans, min(sum[i - 1], i - sum[i - 1]) + min(cnt0, cnt1));
    }
    ans = min(ans, min(sum[n - 1], n - sum[n - 1]));
    cout << ans << '
';
}
 
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    cout << fixed << setprecision(20);
    int T; cin >> T; while(T--)
    run();
    return 0;
}

C. Game On Leaves

首先(x)为叶子结点(注意(n=1)的情况)必胜。
否则最后必胜态为如下形式：

具体证明可参见题解或者视频里所说的。
容易发现最终答案只与结点个数的奇偶性有关，所以判断一下就行。

Code

/*
 * Author:  heyuhhh
 * Created Time:  2020/5/31 23:06:05
 */
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#include <functional>
#include <numeric>
#define MP make_pair
#define fi first
#define se second
#define pb push_back
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define Local
#ifdef Local
  #define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
  void err() { std::cout << std::endl; }
  template<typename T, typename...Args>
  void err(T a, Args...args) { std::cout << a << ' '; err(args...); }
  template <template<typename...> class T, typename t, typename... A> 
  void err(const T <t> &arg, const A&... args) {
  for (auto &v : arg) std::cout << v << ' '; err(args...); }
#else
  #define dbg(...)
#endif
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 1e5 + 5;
 
void run() {
    int n, x; cin >> n >> x; --x;
    vector <int> d(n);
    for (int i = 1; i < n; i++) {
        int u, v; cin >> u >> v;
        --u, --v;
        ++d[u], ++d[v];
    }
    if (d[x] <= 1) {
        cout << "Ayush" << '
';
        return;   
    }
    if (n & 1) cout << "Ashish" << '
';
    else cout << "Ayush" << '
';
}
 
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    cout << fixed << setprecision(20);
    int T; cin >> T; while(T--)
    run();
    return 0;
}

D. Guess The Maximums

交互题。
存在一种刚好用(12)次的方法，具体方法可以看看视频题解，注意第一次询问我们直接对两个集合进行询问，同样能找到最大值，并且还可以在分治的时候省略一层。
细节见代码：

Code

/*
 * Author:  heyuhhh
 * Created Time:  2020/5/31 23:56:07
 */
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#include <functional>
#include <numeric>
#include <random>
#define MP make_pair
#define fi first
#define se second
#define pb push_back
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define Local
#ifdef Local
  #define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
  void err() { std::cout << std::endl; }
  template<typename T, typename...Args>
  void err(T a, Args...args) { std::cout << a << ' '; err(args...); }
  template <template<typename...> class T, typename t, typename... A> 
  void err(const T <t> &arg, const A&... args) {
  for (auto &v : arg) std::cout << v << ' '; err(args...); }
#else
  #define dbg(...)
#endif
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 1e5 + 5;
mt19937 rnd(time(NULL));
 
int n, k;
 
int query2(vector <int>& a) {
    cout << "? " << sz(a) << '
';
    for (auto it : a) cout << it << ' ';
    cout << endl;
    int x; cin >> x;
    return x;   
}
 
int query(vector <int> a) {
    sort(all(a));
    vector <int> b;
    int j = 0;
    for (int i = 1; i <= n; i++) {
        if (j >= sz(a) || i != a[j]) {
            b.push_back(i);
        } else ++j;
    }
    return query2(b);
}
 
void run() {
    cin >> n >> k;
    vector <int> ans(k);
    vector <vector<int>> s(k);
    for (int i = 0; i < k; i++) {
        int c; cin >> c;
        s[i].resize(c);
        for (int j = 0; j < c; j++) {
            cin >> s[i][j];
        }
        sort(all(s[i]));
    }
    int Max;
    int m = (k - 1) / 2;
    vector <int> a, b;
    for (int i = 0; i <= m; i++) {
        for (auto it : s[i]) {
            a.push_back(it);
        }   
    }
    for (int i = m + 1; i < k; i++) {
        for (auto it : s[i]) {
            b.push_back(it);
        }
    }
    int Max1 = query(a), Max2 = query(b);
    Max = max(Max1, Max2);
    if (Max1 == Max2) {
        for (int i = 0; i < k; i++) {
            ans[i] = Max1;
        }
    } else {
        function <int(int, int)> solve;
        solve = [&] (int l, int r) -> int {
            if (l == r) return l;
            int m = (l + r) >> 1;
            vector <int> a;
            for (int i = l; i <= m; i++) {
                for (auto it : s[i]) {
                    a.push_back(it);
                }
            }
            int now = query(a);
            if (now == Max) return solve(m + 1, r);
            return solve(l, m);
        };
        int t;
        if (Max1 < Max2) t = solve(0, m);
        else t = solve(m + 1, k - 1);
        for (int i = 0; i < k; i++) {
            if (i != t) ans[i] = Max;
            else ans[i] = query(s[i]);
        }              
    }
    cout << "!";
    for (auto it : ans) cout << ' ' << it;
    cout << endl;
    string ok; cin >> ok;
    if (ok != "Correct") exit(0);
}
 
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    cout << fixed << setprecision(20);
    int T; cin >> T; while(T--)
    run();
    return 0;
}

E. Tree Shuffling

注意到(a[u]=min(a[u],a[fa[u]]))，那么执行完这个操作过后问题将变为一个子树内的问题，与父亲结点无关。
在一个子数内的话直接贪心将儿子结点剩余进行匹配即可。

Code

/*
 * Author:  heyuhhh
 * Created Time:  2020/5/31 23:20:34
 */
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#include <functional>
#include <numeric>
#define MP make_pair
#define fi first
#define se second
#define pb push_back
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define Local
#ifdef Local
  #define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
  void err() { std::cout << std::endl; }
  template<typename T, typename...Args>
  void err(T a, Args...args) { std::cout << a << ' '; err(args...); }
  template <template<typename...> class T, typename t, typename... A> 
  void err(const T <t> &arg, const A&... args) {
  for (auto &v : arg) std::cout << v << ' '; err(args...); }
#else
  #define dbg(...)
#endif
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 2e5 + 5;
 
int n;
vector <int> G[N];
int a[N], b[N], c[N];
int cnt[N][2];
ll ans;
 
void dfs(int u, int fa) {
    if (fa) a[u] = min(a[u], a[fa]);
    int f = -1;
    if (b[u] == 0 && c[u]) f = 0;
    if (b[u] && c[u] == 0) f = 1;
    int c0 = 0, c1 = 0;
    for (auto v : G[u]) if (v != fa) {
        dfs(v, u);
        cnt[u][0] += cnt[v][0];
        cnt[u][1] += cnt[v][1];
        if (cnt[v][0] > cnt[v][1]) c0 += cnt[v][0] - cnt[v][1];
        else c1 += cnt[v][1] - cnt[v][0];
    }
    if (f == 0) ++cnt[u][0], ++c0;
    if (f == 1) ++cnt[u][1], ++c1;
    ans += 2ll * a[u] * min(c0, c1);
}
 
void run() {
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i] >> b[i] >> c[i];
    }
    for (int i = 1; i < n; i++) {
        int u, v; cin >> u >> v;
        G[u].push_back(v);
        G[v].push_back(u);
    }
    dfs(1, 0);
    if (cnt[1][0] != cnt[1][1]) ans = -1;
    cout << ans << '
';
}
 
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    cout << fixed << setprecision(20);
    run();
    return 0;
}

F. Rotating Substrings

很神仙的dp，不好想。
有两种做法，第一种做法可以参见视频题解里面说的，我们要求一个最长的公共子序列，并且剩下的每一个在(s)串中没有匹配的数都只能往前移动。
那么即是对每一个匹配的(s_i,t_j),(i+1 ightarrow n)中(x)字符出现的次数不小于(t+1 ightarrow n)中出现的次数，这样的话最终剩下没有匹配的只能往右边走。那么最终答案即为(n-dp[n][n])。
另外一种做法也很神奇！我们定义(dp[i][j])为(s)串中确定了(i)个字符，(t)串匹配到了(j)时，最少需要插入的数的个数，显然最终答案即为(dp[n][n])。
那么我们将题目中的操作看作两部分，一部分是选出一些数准备插入，另外一部分就是插入。
那么往后移动(j)时其实我们在不断选择一些数出来，然后往后移动(i)时我们在不断插入一些数，代价为(1)，当然如果(s[i]=t[j])的话就不用消耗代价。
细节见代码吧。。

solution1

/*
 * Author:  heyuhhh
 * Created Time:  2020/6/1 13:32:17
 */
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#include <functional>
#include <numeric>
#define MP make_pair
#define fi first
#define se second
#define pb push_back
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define Local
#ifdef Local
  #define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
  void err() { std::cout << std::endl; }
  template<typename T, typename...Args>
  void err(T a, Args...args) { std::cout << a << ' '; err(args...); }
  template <template<typename...> class T, typename t, typename... A> 
  void err(const T <t> &arg, const A&... args) {
  for (auto &v : arg) std::cout << v << ' '; err(args...); }
#else
  #define dbg(...)
#endif
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 2000 + 5;

void run() {
    int n; cin >> n;
    string s, t; cin >> s >> t;
    string ss = s, tt = t;
    sort(all(ss)), sort(all(tt));
    if (ss != tt) {
        cout << -1 << '
';
        return;
    }
    s = '#' + s;
    t = '#' + t;
    vector <vector <int>> dp(n + 1, vector<int>(n + 1));
    vector <vector <int>> suf_s(n + 2, vector<int>(26)), suf_t(n + 2, vector<int>(26));
    for (int i = n; i >= 1; i--) {
        for (int j = 0; j < 26; j++) {
            suf_s[i][j] = suf_s[i + 1][j];
            suf_t[i][j] = suf_t[i + 1][j];
        }
        ++suf_s[i][s[i] - 'a'];
        ++suf_t[i][t[i] - 'a'];
    }
    dp[0][0] = 0;
    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= n; j++) {
            if (i) dp[i][j] = dp[i - 1][j];
            if (j) dp[i][j] = max(dp[i][j], dp[i][j - 1]);
            if (i && j && s[i] == t[j]) {
                bool ok = true;
                for (int k = 0; k < 26; k++) {
                    if (suf_s[i][k] < suf_t[j][k]) ok = false;   
                }
                if (ok) {
                    dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + 1);   
                }
            }
        }   
    }
    cout << n - dp[n][n] << '
';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    cout << fixed << setprecision(20);
    int T; cin >> T; while(T--)
    run();
    return 0;
}

solution1

/*
 * Author:  heyuhhh
 * Created Time:  2020/6/1 21:01:21
 */
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#include <assert.h>
#include <functional>
#include <numeric>
#define MP make_pair
#define fi first
#define se second
#define pb push_back
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define Local
#ifdef Local
  #define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
  void err() { std::cout << std::endl; }
  template<typename T, typename...Args>
  void err(T a, Args...args) { std::cout << a << ' '; err(args...); }
  template <template<typename...> class T, typename t, typename... A> 
  void err(const T <t> &arg, const A&... args) {
  for (auto &v : arg) std::cout << v << ' '; err(args...); }
#else
  #define dbg(...)
#endif
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 1e5 + 5;

void run() {
    int n; cin >> n;
    string s, t; cin >> s >> t;
    string ss = s, tt = t;
    sort(all(ss)), sort(all(tt));
    if (ss != tt) {
        cout << -1 << '
';
        return;
    }
    s = '#' + s;
    t = '#' + t;
    vector <vector <int>> dp(n + 1, vector<int>(n + 1, INF));   
    vector <vector <int>> suf_s(n + 2, vector<int>(26)), suf_t(n + 2, vector<int>(26));
    for (int i = n; i >= 1; i--) {
        for (int j = 0; j < 26; j++) {
            suf_s[i][j] = suf_s[i + 1][j];
            suf_t[i][j] = suf_t[i + 1][j];
        }
        ++suf_s[i][s[i] - 'a'];
        ++suf_t[i][t[i] - 'a'];
    }
    for (int i = 0; i <= n; i++) {
        dp[0][i] = 0;
    }
    for (int i = 1; i <= n; i++) {
        for (int j = i; j <= n; j++) {
            if (s[i] == t[j]) {
                dp[i][j] = dp[i - 1][j - 1];
            }
            dp[i][j] = min(dp[i][j], dp[i - 1][j] + 1);
            if (suf_s[i + 1][t[j] - 'a'] > suf_t[j + 1][t[j] - 'a']) {
                dp[i][j] = min(dp[i][j], dp[i][j - 1]);
            }
        }
    }
    cout << dp[n][n] << '
';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    cout << fixed << setprecision(20);
    int T; cin >> T; while(T--)
    run();
    return 0;
}