2022 China Collegiate Programming Contest (CCPC) Mianyang Onsite （2022CCPC绵阳）ACGHM

https://codeforces.com/gym/104065
昨天女队vp了一下，赛时4题223罚时
A是一个dp，学妹已经写的差不多了，就是转移方向错了（给点时间应该能d出来）牛的牛的。我就看了点签到，又是作为蟑螂乱爬的一天

A. Ban or Pick, What's the Trick

记忆化搜索。
状态：\(dp_{i, j, k}\) 表示进行到第 \(i\) 轮，\(A\) 选了 \(j\) 个， \(B\) 选了 \(k\) 个的 \(s_A-s_B\) 的值，然后二者都是从大往小来选（贪心），所以当前选的数也可以通过状态来计算出他所在的下标。

#include <bits/stdc++.h>

using namespace std;
const int N = 2e5 + 5;
int n, m, a[N], b[N], f[N][11][11];
bool vis[N][11][11];

int dfs (int x, int ca, int cb) {
    if (x >= 2 * n) return 0;
    if (vis[x][ca][cb])   return f[x][ca][cb];
    vis[x][ca][cb] = true;
    int aa = x / 2 - cb + ca + 1, bb = (x + 1) / 2 - ca + cb + 1, &t = f[x][ca][cb];
    t = dfs (x + 1, ca, cb); //不选
    if (x % 2 == 0) { //A
        if (aa <= n && ca < m)  t = max (t, dfs (x + 1, ca + 1, cb) + a[aa]);
    }   
    else { //B
        if (bb <= n && cb < m)  t = min (t, dfs (x + 1, ca, cb + 1) - b[bb]);
    }
    return t;
}

int main () {
    scanf ("%d%d", &n, &m);
    for (int i = 1; i <= n; i++)    scanf ("%d", &a[i]);
    for (int i = 1; i <= n; i++)    scanf ("%d", &b[i]);
    sort (a + 1, a + n + 1, greater<int>());
    sort (b + 1, b + n + 1, greater<int>());
    printf ("%d\n", dfs (0, 0, 0));
}

C. Catch You Catch Me

签到。答案为所有深度为1的点的子树最大深度之和。

#include <bits/stdc++.h>
#define ll long long

using namespace std;
const int N = 1e5 + 5, M = N * 2;
int n, h[N], e[M], ne[M], idx;
ll dep[N], ans, tt;
bool d[N];

void add (int a, int b) {
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

int dfs (int u, int fa) {
    dep[u] = dep[fa] + 1, tt = max (tt, dep[u]);
    for (int i = h[u]; ~i; i = ne[i]) {
        int j = e[i];
        if (j == fa)    continue;
        dfs (j, u);
    }
    return tt;
}

int main () {
    memset (h, -1, sizeof h);
    scanf ("%d", &n);
    for (int i = 1; i <= n; i++) {
        int a, b;
        scanf ("%d%d", &a, &b);
        add (a, b), add (b, a);
        if (a == 1) d[b] = true;
        if (b == 1) d[a] = true;
    }

    for (int i = 1; i <= n; i++) {
        if (d[i])    tt = 1, ans += dfs (i, 1);// cout << i << ' ' << tt << endl;
    }
    printf ("%d\n", ans);
}

G. Let Them Eat Cake

模拟。因为每次至少会删掉一半的数，所以最多进行log次
暴力模拟即可

#include <bits/stdc++.h>
#define ll long long

using namespace std;
const int N = 1e5 + 5;
int n, a[N], ans;
bool vis[N];

int main () {
    scanf ("%d", &n);
    vector<int> v;
    for (int i = 1; i <= n; i++)    scanf ("%d", &a[i]), v.push_back (a[i]);
    while (1) {
        vector<int> tt;
        n = v.size ();
        if (n == 1) break;
        if (v[0] > v[1])    tt.push_back (v[0]);
        for (int i = 1; i < n - 1; i++) {
            if (v[i] > v[i+1] && v[i] > v[i-1]) tt.push_back (v[i]);
        }
        if (v[n-1] > v[n-2])    tt.push_back (v[n-1]);
        v.clear ();
        for (auto i : tt)   v.push_back (i);// cout << i << ' ';
        //cout << endl;
        ans++;
        if (v.size () == 1)   break;
    }
    
    printf ("%d\n", ans);
}

H - Life is Hard and Undecidable, but...

构造一条斜线即可，然而我们想了好久qaq

#include <bits/stdc++.h>
#define ll long long

using namespace std;

int main () {
    int n;
    cin >> n;
    cout << 2 * n - 1 << endl;
    for (int i = 1; i < 2 * n; i++) cout << i << ' ' << i << endl;
}

M. Rock-Paper-Scissors Pyramid

栈模拟，维护栈内始终是 a 赢 b 的关系
最终答案就是栈底

#include <bits/stdc++.h>
#define ll long long

using namespace std;

bool check (char a, char b) { //a输给b
    if (a == b) return true;
    if (a == 'S')   return b == 'R';
    if (a == 'P')   return b == 'S';
    return b == 'P';
}

void solve () {
    string s;
    cin >> s;
    stack <char> stk;
    stk.push (s[0]);
    for (int i = 1; i < s.size (); i++) {
        while (stk.size () && check (stk.top (), s[i])) stk.pop ();
        stk.push (s[i]);
    }

    while (stk.size () > 1) stk.pop ();
    cout << stk.top () << endl;
}

int main () {
    int t;
    cin >> t;
    while (t--) solve ();
}

剩下两题看半天题解看不懂5555我好菜