AtCoder Beginner Contest 302

Posted 2023-05-25 PHarr

A - Attack

#include <bits/stdc++.h>

using namespace std;

#define int long long

int32_t main() 
    int a , b;
    cin >> a >> b;
    cout << (a+b-1) / b << "\\n";
    return 0;

B - Find snuke

直接枚举八个方向，暴力判断。

#include <bits/stdc++.h>

using namespace std;

#define int long long

const int dx[] = 0, 0, 1, -1, 1, 1, -1, -1;
const int dy[] = 1, -1, 0, 0, 1, -1, 1, -1;
const string p = "snuke";

int32_t main() 
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    int n, m;
    cin >> n >> m;
    vector<string> g(n);
    for (auto &i: g) cin >> i;
    for ( auto i : g )

    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            for (int l = 0, f; l < 8; l++) 

                if (i + dx[l] * 4 < 0 || i + dx[l] * 4 >= n || j + dy[l] * 4 < 0 || j + dy[l] * 4 >= m) continue;
                f = 1;

                for (int x = i, y = j, t = 0; f && t < 5; t++, x += dx[l], y += dy[l])
                    if (g[x][y] != p[t]) f = 0;
                if (f == 0) continue;
                for (int x = i, y = j, t = 0; f && t < 5; t++, x += dx[l], y += dy[l])
                    cout << x+1 << " " << y+1 << "\\n";
                return 0;
            
    return 0;

C - Almost Equal

枚举全排列，然后暴力判断

#include <bits/stdc++.h>

using namespace std;

#define int long long

int32_t main() 
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    int n, m;
    cin >> n >> m;
    vector<string> g(n);
    for (auto &i: g) cin >> i;
    sort(g.begin(), g.end());
    do
        int f = 1;
        for( int i = 1 , cnt ; f && i < n ; i ++ )
            cnt = 0;
            for( int j = 0 ; j < m ; j ++ )
                cnt += ( g[i][j] != g[i-1][j] );
            if( cnt != 1 ) f = 0;
        
        if( f == 0 ) continue;
        cout << "Yes\\n";
        return 0;
    while(next_permutation(g.begin(),g.end()));
    cout << "No\\n";
    return 0;

D - Impartial Gift

枚举一个，二分另一个即可。

#include <bits/stdc++.h>

using namespace std;

#define int long long

int32_t main() 
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    int n, m, d;
    cin >> n >> m >> d;
    vector<int> a(n), b(m);
    for (auto &i: a) cin >> i;
    for (auto &i: b) cin >> i;
    sort(a.begin(),a.end()) , sort(b.begin(),b.end());
    int res = -1;
    for( auto i : a )
        auto j = lower_bound(b.begin(),b.end(),i+d);
        while( j != b.begin() && (j == b.end() || (*j) > i + d) ) j = prev(j);
        if( abs((*j) - i ) > d ) continue;
        res = max( res , (*j) + i );
    
    cout << res << "\\n";
    return 0;

E - Isolation

其实直接维护一下每个点连边情况，动态的统计答案就好了。

#include <bits/stdc++.h>

using namespace std;

#define int long long

int32_t main() 
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    int n, q, cnt;
    cin >> n >> q, cnt = n;
    vector<set<int>> e(n + 1);
    for (int op, u, v; q; q--) 
        cin >> op;
        if (op == 1) 
            cin >> u >> v;
            if (e[u].empty()) cnt--;
            if (e[v].empty()) cnt--;
            e[u].insert(v), e[v].insert(u);

         else 
            cin >> u;
            for (auto v: e[u]) 
                e[v].erase(u);
                if (e[v].empty()) cnt++;
            
            if (!e[u].empty()) e[u].clear(), cnt++;
        
        cout << cnt << "\\n";
    
    return 0;

F - Merge Set

把每一个集合当作一个虚点，把集合内的元素与集合连边，集合的点权为 1，元素的点权为 0。这样从 1 开始求一遍最短路，dis[i]就表示了把i和 1 联通所需的最少集合数目。那么答案就是dis[m]-1。

#include <bits/stdc++.h>

using namespace std;

#define int long long


int32_t main() 
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    int n, m;
    cin >> n >> m;
    vector<vector<int>> e(n + m + 2);
    for (int i = 1, s; i <= n; i++) 
        cin >> s;
        for (int y; s; s--) 
            cin >> y;
            e[y].push_back(m + i), e[m + i].push_back(y);
        
    
    vector<int> dis(n + m + 2, INT_MAX);
    vector<bool> vis(n + m + 2, 0);
    dis[1] = 0;
    priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> q;
    q.emplace(0, 1);
    while (!q.empty()) 
        auto [d, u] = q.top();

        q.pop();
        if (vis[u]) continue;
        vis[u] = 1;
        for (auto v: e[u]) 
            if (vis[v] || dis[v] <= d + (v <= m)) continue;
            dis[v] = d + (v <= m), q.emplace(dis[v], v);
        
    

    if (dis[m] == INT_MAX) cout << -1;
    else cout << dis[m] - 1;

    return 0;

G - Sort from 1 to 4

置换环可以得到数组排序（可以指定排序方式）所需交换的最小次数。其的思想是：将每个节点指向其排序后应该存放的位置，最终首位相接形成一个环，那么数组排序所需的最小交换次数为数组长度-环的数量。

因为这里数据范围只有\\([1,4]\\)，所以置换环的大小也就是\\(2,3,4\\)，我们可以直接暴力的枚举置换环，然后计算置换环的数量即可。

#include <bits/stdc++.h>

using namespace std;

#define int long long

int cnt[5][5];

int32_t main() 
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    int n, res = 0;
    cin >> n, res = 0;
    vector<int> a(n);
    for (auto &i: a) cin >> i;
    auto b = a;
    sort(b.begin(), b.end());

    for (int i = 0; i < n; i++)
        if (a[i] != b[i]) cnt[a[i]][b[i]]++;


    for( int i = 1 ; i <= 4 ; i ++ )
        for( int j = i + 1 , cur ; j <= 4 ; j ++ )
            cur = min( cnt[i][j] , cnt[j][i] );
            res += cur , cnt[i][j] -= cur , cnt[j][i] -= cur;
        
    
    for( int i = 1 ; i <= 4 ; i ++ )
        for( int j = 1 ; j <= 4 ; j ++ )
            for( int k = 1 , cur ; k <= 4 ; k ++ )
                if( i == j || i == k || j == k ) continue;
                cur = min( cnt[i][j] , cnt[j][k] , cnt[k][i]  );
                cnt[i][j] -= cur , cnt[j][k] -= cur , cnt[k][i] -= cur;
                res += cur * 2;
            

    for( int i = 2 ; i <= 4 ; i ++ )
        res += cnt[1][i] * 3;
    cout << res << "\\n";
    return 0;

Atcoder Beginner Contest 257

A - A to Z String 2 代码： void solve(int Case) int n; string s = " "; cin >> n; int k; cin >> k; for (int i = 'A'; i <= 'Z'; i++) for (int j = 1; j