Codeforces Round #670 (Div. 2)

题目传送门

A. Subset Mex

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, a, b) for (register int i = a; i <= b; i++)
 
int n;
int a[110], cnt[110];
inline void solve(int T) {
 
    cin >> n;
    int ans = 0;
    rep(i, 0, 105) cnt[i] = 0;
    rep(i, 1, n) cin >> a[i], cnt[a[i]]++;
    rep(i, 0, 105) if(cnt[i]) {
        cnt[i]--;
    } else {
        ans += i;
        break;
    }
    rep(i, 0, 105) if(cnt[i]) {
        cnt[i]--;
    } else {
        ans += i;
        break;
    }
    cout << ans << endl;
    
 
    
}
int main()
{
    // ios_base::sync_with_stdio(0);
    // cin.tie(0);
    // cout.tie(0);
 
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
 
    int T = 1;
    cin >> T;
    rep(i, 1, T) solve(i);
    
    // system("pause");
}

B. Maximum Product

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, a, b) for (register int i = a; i <= b; i++)
 
ll n;
ll a[100010];
 
inline void solve(int T) {
 
    cin >> n;
    rep(i, 1, n) cin >> a[i];
    sort(a + 1, a + n + 1);
    ll ans = a[1] * a[2] * a[3] * a[4] * a[5];
    ans = max(ans, a[1] * a[2] * a[3] * a[4] * a[n]);
    ans = max(ans, a[1] * a[2] * a[3] * a[n - 1] * a[n]);
    ans = max(ans, a[1] * a[2] * a[n - 2] * a[n - 1] * a[n]);
    ans = max(ans, a[1] * a[n - 3] * a[n - 2] * a[n - 1] * a[n]);
    ans = max(ans, a[n - 4] * a[n - 3] * a[n - 2] * a[n - 1] * a[n]);
    cout << ans<< endl;
 
    
}
int main()
{
    // ios_base::sync_with_stdio(0);
    // cin.tie(0);
    // cout.tie(0);
 
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
 
    int T = 1;
    cin >> T;
    rep(i, 1, T) solve(i);
    
    // system("pause");
}

C. Link Cut Centroids

主要是找重心，如果有两个就把其中一个作为根，找到最大的连通块取一个点连向根

解释的不是很清楚，具体看代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, a, b) for (register int i = a; i <= b; i++)
 
int n;
vector<int> son[100010];
int sz[100010], max_part[100010];
int vis[100010];
int ma, cnt, point;
 
void dfs(int u, int fa) {
    sz[u] = 1;
    max_part[u] = 0;
    for(vector<int>::iterator it = son[u].begin(); it != son[u].end(); it++) {
        int v = *it;
        if(v == fa) continue;
        dfs(v, u);
        sz[u] += sz[v];
        max_part[u] = max(max_part[u], sz[v]);
    }
    max_part[u] = max(max_part[u], n - sz[u]);
    if(max_part[u] < ma) {
        cnt = 1;
        ma = max_part[u];
        point = u;
    }
    else if(max_part[u] == ma) cnt++;
}
bool cmp(int ff, int gg) {return sz[ff] > sz[gg];}
 
inline void solve(int T) {
 
    cin >> n;
    rep(i, 1, n) son[i].clear();
    ma = n * 2;
    cnt = point = 0;
    int x, y;
    rep(i, 1, n - 1) {
        cin >> x >> y;
        son[x].push_back(y);
        son[y].push_back(x);
    }
    dfs(1, 0);
    // cout << max_part[point] << " " << point << " " << cnt << endl;
    if(cnt == 1) cout << x << " " << y << endl << x << " " << y << endl;
    else {
        dfs(point, 0);
        int u = point, v;
        while(son[u].size() > 1) {
            sort(son[u].begin(), son[u].end(), cmp);
            v = u;
            if(u == point) u = son[u][0];
            else u = son[u][1];
        }
        cout << u << " " << v << endl << u << " " << point << endl;
    }
    
}
int main()
{
    // ios_base::sync_with_stdio(0);
    // cin.tie(0);
    // cout.tie(0);
 
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
 
    int T = 1;
    cin >> T;
    rep(i, 1, T) solve(i);
    
    // system("pause");
}