The 2023 Guangdong Provincial Collegiate Programming Contest

A - 算法竞赛

#include <bits/stdc++.h>
using namespace std;
#define int long long

void solve(){
    int st , n , ed;
    cin >> st >> n;
    map<int,int> cnt;
    for( int i = 1 , x ; i <= n ; i ++ ){
        cin >> x;
        cnt[x] ++;
    }
    cin >> ed;
    int res = 0;
    for( int i = st ; i <= ed ; i ++ ){
        if( cnt[i] ) continue;
        res ++;
    }
    cout << res << "\n";
    return ;
}

int32_t main() {
    int t;
    cin >> t;
    for( ; t ; t -- ){
        solve();
    }
    return 0;
}

B - 基站建设

\(f(i)\)表示前\(i\)个基站\(i\)必选的情况下的最小成本\(f(i) =\min f(j) + a_i\)

注意\([j+1,i-1]\)中不能有任何一个完整区间，所以我们要计算出\(lim(i)\)表示\([lim(i),i]\)中没有完整区间的最小值。那么\(j\ge lim(i-1)-1\)

然后再用单调队列优化一下dp

#include <bits/stdc++.h>

using namespace std;

#define int long long

void solve() {
    int n;
    cin >> n;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; i++) cin >> a[i];
    // 在 n+1 添加一个花费为 0 的基站，并添加一个 [n+1,n+1] 的需求区间这样 f[n+1] 就是答案
    a.push_back(0), n++;
    int m;
    cin >> m;
    vector<vector<int>> b(n+1);
    // 里面的负数 -j 表示有一个需求区间是 [i, j]
    // 里面是正数 j 表示有一个需求区间是 [j, i]
    for (int i = 1, l, r; i <= m; i++)
        cin >> l >> r, b[l].push_back(-r), b[r].push_back(l);
    b[n].push_back( -n) ,b[n].push_back(n);

    vector<int> lim(n+1);
    for( int l = 1 , r = 1 , cnt = 0; l <= n ; r ++ ){
        for( auto x : b[r] )
            if( x > 0 && x >= l ) cnt ++;
        while( cnt > 0 && l <= r ){
            for( auto x : b[l] )
                if( x < 0 && -x <= r ) cnt --;
            l ++;
        }
        lim[r] = l;
    }
    vector<int> f( n+1 );
    deque<int> q;
    f[1] = a[1] , q.push_back(0) , q.push_back(1);
    for( int r = 2 , l ; r <= n ; r ++ ){
        l = lim[r-1] - 1;
        while( q.front() < l ) q.pop_front();
        f[r] = f[ q.front() ] + a[r];
        while( !q.empty() && f[ q.back() ] >= f[r] ) q.pop_back();
        q.push_back(r);
    }
    cout << f[n] << "\n";
}

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int t;
    cin >> t;
    for (; t; t--)
        solve();
    return 0;
}

C - 市场交易

按照价格排序后贪心购买

#include <bits/stdc++.h>
using namespace std;

#define int long long

void solve(){
    int n;
    cin >> n;
    vector<pair<int,int>> a(n);
    for( auto & [x , y ] : a )
        cin >> x >> y;
    sort( a.begin(), a.end() );

    int res = 0;
    for( int l = 0 , r = n-1 , t ; l < r ; ){
        t = min( a[l].second , a[r].second );
        res += (a[r].first - a[l].first) * t;
        a[l].second -= t , a[r].second -= t;
        if( a[l].second == 0 ) l ++;
        if( a[r].second == 0 ) r --;
    }
    cout << res << "\n";
    return ;
}

int32_t main() {
    ios::sync_with_stdio(false) , cin.tie(nullptr) , cout.tie(nullptr);
    int t;
    cin >> t;
    for( ; t ; t -- ){
        solve();
    }
    return 0;
}

D - 新居规划

把人按照\(a_i-b_i\)从小到大排序，排序后越靠前的人越喜欢独居，越靠后的人越喜欢群居，然后分别计算前缀独居贡献和和后缀群居贡献和。最后枚举出独居的人数即可。

#include <bits/stdc++.h>
using namespace std;

#define int long long

void solve(){
    int n , m;
    cin >> n >> m;

    vector<pair<int,int>> a(n);
    for( auto &[x,y] : a )
        cin >> x >> y;
    sort( a.begin(), a.end() , [](pair<int,int> x , pair<int,int>y){
        return x.first-x.second < y.first - y.second;
    } );
    vector<int> p(n+1) , q(n+1);

    for( int i = 1 ; i <= n ; i ++ )
        p[i] = p[i-1] + a[i-1].second;

    q[n] = a[n-1].first;
    for( int i = n-1 ; i >= 1 ; i -- )
        q[i] = q[i+1] + a[i-1].first;

    reverse(q.begin()+1, q.end());

    int res = 0;
    for( int i = 0 , j = n ; i <= n ; i ++ , j -- ){
        if( i * 2 + j > m ) break;
        if( j < 2 ) continue;
        res = max( res , p[i] + q[j] );
    }
    if( 2 * n - 1 <= m ) res = max( res , p[n] );
    cout << res << "\n";


    return ;
}

int32_t main() {
    ios::sync_with_stdio(false) , cin.tie(nullptr) , cout.tie(nullptr);
    int t;
    cin >> t;
    for( ; t ; t -- ){
        solve();
    }
    return 0;
}

E - 新杯质问

把所有的字符串都插入到 Tire 上，然后贪心的选择即可，首先给每个分支都任一选择一个，这样的话 LCP不会变化。如果不能满足选择的需求的话，就按照字典序贪心的选择，最后一个选到的就是 LCP 新增的一位。

#include <bits/stdc++.h>

using namespace std;

#define int long long

vector<int> val;

struct node {
    vector<int> to;
    int g, cnt;
    node() {
        to = vector<int>(26, -1);
        cnt = 0, g = -1;
    }
};

vector<node> tire;

string res;

void dfs(int x) {
    if (tire[x].cnt) val[x] = tire[x].cnt;
    for (auto y: tire[x].to) {
        if (y == -1) continue;
        dfs(y);
        val[x] += val[y];
    }
    return;
}

void calc(int x, int k) {
    if (tire[x].g != -1) res += (char) (tire[x].g + 'a');
    if( tire[x].cnt ) k -= tire[x].cnt;
    if( k <= 1 ) return ;
    int cnt = 26;
    for (auto y: tire[x].to)
        cnt -= (y == -1);
    if (cnt >= k) return;
    for (auto y: tire[x].to) {
        if (y == -1) continue;
        cnt--;
        if (cnt + val[y] < k) k -= val[y];
        else {
            k -= cnt;
            calc(y, k);
            return;
        }
    }
}

void solve() {
    int n, m;
    cin >> n >> m;
    tire = vector<node>(1, node());
    for (int i = 1, t; i <= n; i++) {
        string s;
        cin >> s;
        t = 0;
        for (int i = 0, j; i < s.size(); i++) {
            j = s[i] - 'a';
            if (tire[t].to[j] == -1) {
                tire[t].to[j] = tire.size();
                tire.push_back(node());
                tire.back().g = j;
            }
            t = tire[t].to[j];
        }
        tire[t].cnt ++;
    }
    val = vector<int>(tire.size());
    dfs(0);
    res = "";
    calc(0, m);
    if (res == "") res = "EMPTY";
    cout << res << "\n";
}

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    int t;
    cin >> t;
    for (; t; t--) {
        solve();
    }
    return 0;
}

I - 路径规划

二分答案，二分\(mex\)的值，则\([0,mex-1]\)都应该在序列中，这样这些点的下标按照\(x\)升序排列后，\(y\)也应该是升序排序。所以\(O(n\log n )\)的校验即可

#include <bits/stdc++.h>

using namespace std;


void solve() {
    int n, m;
    cin >> n >> m;
    vector<pair<int, int>> a(n * m);
    for( int i = 1 ; i <= n ; i ++ )
        for( int x , j = 1 ; j <= m ; j ++ )
            cin >> x , a[x] = make_pair( i , j );

    auto check =[a]( int x ){
        if( x < 0 ) return true;
        auto b = vector<pair<int,int>>( a.begin() , a.begin() + x );
        sort( b.begin() , b.end() );
        for( int i = 1 ; i < b.size() ; i ++ )
            if( b[i].second < b[i-1].second ) return false;
        return true;
    };

    int l = 0 , r = n*m , mid ,res = 0;
    while( l <= r ){
        mid = ( l + r ) >> 1;
        if( check( mid ) ) res = mid , l = mid + 1;
        else r = mid - 1;
    }
    cout << res << "\n";
    return;
}

int main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    int t;
    cin >> t;
    for (; t; t--) {
        solve();
    }
    return 0;
}

K - 独立钻石

直接 dfs，暴力的枚举棋子和方向即可，\(O(6!\times 4)\)

#include <bits/stdc++.h>

using namespace std;

int res, n, m;

const int dx[] = {0, 0, 1, -1};
const int dy[] = {1, -1, 0, 0};

void dfs(int k, vector<vector<int>> g) {
    res = min(res, k);
    if( k == 1 ) return ;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (g[i][j] == 0) continue;
            for (int l = 0, ax, ay, bx, by; l < 4; l++) {
                ax = i + dx[l], ay = j + dy[l];
                if (ax < 0 || ay < 0 || ax >= n || ay >= m || g[ax][ay] == 0) continue;
                bx = ax + dx[l], by = ay + dy[l];
                if (bx < 0 || by < 0 || bx >= n || by >= m || g[bx][by] == 1) continue;
                g[i][j] = 0, g[ax][ay] = 0, g[bx][by] = 1;
                dfs(k-1, g);
                g[i][j] = 1, g[ax][ay] = 1, g[bx][by] = 0;
            }
        }
    }
}

void solve() {
    int k;
    cin >> n >> m >> k;
    vector<vector<int>> g(n, vector<int>(m, 0));
    res = k;
    for (int x, y; k; k--)
        cin >> x >> y , x -- , y -- ,  g[x][y] = 1;
    dfs(res, g);
    cout << res << "\n";
    return;
}

int main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    int t;
    cin >> t;
    for (; t; t--) {
        solve();
    }
    return 0;
}