L1-1 谢谢卡尔！

print("谢谢卡尔！\\(>_<)/")

L1-2 现在是，幻想时间！

#include <bits/stdc++.h>

using namespace std;

int main(){
    double a , b;
    cin >> a >> b;
    printf("%.3lf" , a / b );
}

L1-3 你是来陪可莉炸鱼的吗?

#include <bits/stdc++.h>

using namespace std;

int main() {
    int n, res = 0;
    cin >> n;
    for (int x; n; n--) {
        cin >> x;
        if (x < 100) res += 1;
        else if (x < 200) res += 2;
        else if (x < 300) res += 5;
        else if (x < 400) res += 10;
        else res += 15;
    }
    cout << res;
}

L1-4 扫雷游戏

#include <bits/stdc++.h>

using namespace std;

int f[15][15], n, m;
const int dx[] = {0, 0, 1, -1, 1, 1, -1, -1};
const int dy[] = {1, -1, 0, 0, 1, -1, 1, -1};

int main() {
    cin >> n >> m;
    string s;
    for (int i = 1; i <= n; i++) {
        cin >> s;
        for (int j = 1; j <= m; j++) {
            if (s[j - 1] == '.') continue;
            f[i][j] = INT_MIN;
            for (int k = 0; k < 8; k++)
                f[i + dx[k]][j + dy[k]]++;
        }
    }
    for( int i = 1 ; i <= n ; i ++ ){
        for( int j = 1 ; j <= m ; j ++ ){
            if( f[i][j] == 0 ) cout << ".";
            else if( f[i][j] < 0 ) cout << "*";
            else cout << f[i][j];
        }
        cout << "\n";
    }
}

L1-5 史莱姆

#include <bits/stdc++.h>

using namespace std;

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

int main() {
    int n = read();
    vector<char> b;
    string s;
    cin >> s;
    for( auto i : s ){
        if( b.empty() ) b.push_back(i);
        else if( i != b.back() ) b.push_back(i);
    }
    cout << b.size() << "\n";
    for( auto i : b ) cout << i;
}

L1-6 加密通信

#include <bits/stdc++.h>

using namespace std;

int main() {
    int n, k;
    string s;
    cin >> n >> k >> s;
    vector<int> key(26);
    for (auto &i: key) cin >> i;
    k %= n , k = n - k;
    for (int i = k; i < n; i++)
        cout << key[s[i] - 'a'];
    for (int i = 0; i < k; i++)
        cout << key[s[i] - 'a'];
}

L1-7 字符操作

#include <bits/stdc++.h>

using namespace std;

int main() {
    int n;
    string s , v[2];
    int a = 0 , b = 1;
    cin >> n >> s;
    v[0] = s.substr(0,n) , v[1] = s.substr(n,n);
    int m; cin >> m;
    for( int op , x , y , p , q ; m ; m -- ){
        cin >> op >> x >> y;
        if( op == 2 ) swap( a , b );
        else {
            x -= 1 , y -= 1;
            if( x < n ) p = a;
            else p = b , x -= n;
            if( y < n ) q = a;
            else q = b , y -= n;
            swap( v[p][x] , v[q][y] );
        }
    }
    cout << v[a] << v[b];

}

L1-8 vivo50！

#include <bits/stdc++.h>

using namespace std;

struct Stu{
    string name;
    double gpa , all;
    int de , zhi;

    Stu( string name , int gpa , int de , int zhi )
        :name(name) , gpa(gpa) , de(de) , zhi(zhi){
        all = double( gpa + zhi ) * 0.7 + 0.3 * de;
    }

    bool operator < ( Stu b ){
        if( all != b.all ) return all > b.all;
        return name < b.name;
    }

};

int main() {
    int n , k;
    cin >> n >> k;
    vector<Stu> s;
    string name;
    double gpa;
    int de , zhi;
    for( int i = 1 , g ; i <= n ; i ++ ){
        cin >> name >> gpa >> de >> zhi >> g;
        if( g == 0 ) continue;
        s.emplace_back( name , gpa * 10 + 50 , min( de + 70 , 100 ) , zhi );
    }
    sort( s.begin(), s.end() );
    for( int i = 0 , rk ; i < s.size() ; i ++ ){
        if( i != 0 && s[i].all == s[i-1].all ) rk = rk;
        else rk = i + 1;
        if( rk > k ) break;
        cout << rk << " " << s[i].name;
        printf(" %.1lf\n" , s[i].all );
    }
}

L2-1 游戏圈

就用并查集维护一下就好了

#include <bits/stdc++.h>

using namespace std;

class dsu{
private:
    vector<int> fa;
public:
    dsu( int n = 1 ){
        fa = vector<int>( n+1 , -1 ) , fa[0] = 0;
    }
    int getfa( int x ){
        if( fa[x] < 0 ) return x;
        return fa[x] = getfa( fa[x] );
    }
    void merge( int x , int y ){
        x = getfa(x) , y = getfa(y);
        if( x == y ) return ;
        if( fa[x] > fa[y] ) swap( x , y );
        fa[x] += fa[y] , fa[y] = x;
    }
    bool check( int x , int y ){
        x = getfa(x) , y = getfa(y);
        return ( x == y );
    }
};

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}
int main() {
    int n = read() , m = read() , q = read();
    dsu t(n);
    for( int u , v ; m ; m -- ){
        u = read() , v = read();
        t.merge( u , v );
    }
    for( int x , y ; q ; q -- ){
        x = read() , y = read();
        if( t.check(x,y) ) printf("yes\n");
        else printf("no\n");
    }
    int cnt = 0;
    for( int i = 1 ; i <= n ; i ++ ){
        if( t.getfa(i) == i ) cnt ++;
    }
    cout << cnt << "\n";
}

如果不会并查集的话当然也可以用搜索搞一搞

#include<bits/stdc++.h>

using namespace std;

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

int32_t main() {
    int n = read() , m = read() , q = read() , cnt = 0;
    vector<vector<int>> e(n+1);
    for( int a , b ; m ; m -- )
        a = read() , b = read() , e[a].push_back(b) , e[b].push_back(a);
    vector<int> vis( n+1 );
    for( int i = 1 ; i <= n ; i ++ ){
        if( vis[i] ) continue;
        cnt ++;
        queue<int> q;
        q.push(i);
        while( !q.empty() ){
            int u = q.front(); q.pop();
            if( vis[u] ) continue;
            vis[u] = i;
            for( auto v : e[u] ){
                if( vis[v] ) continue;
                q.push(v);
            }
        }
    }
    for( int a , b ; q ; q -- ){
        a = read() , b = read();
        if( vis[a] != vis[b] ) printf("no\n");
        else printf("yes\n");
    }
    printf("%d\n" , cnt );
    return 0;
}

L2-2 组套题

模拟题，没什么算法，注意细节就好了。然后就是如何处理读入和输入可以采用格式化的形式，比较方便。

#include <bits/stdc++.h>

using namespace std;

int32_t main() {
    int n;
    scanf("%d" , &n );
    vector<int> a(11);
    for (int i = 1; i <= 10; i++) scanf("%d" , &a[i] );
    vector<tuple<int, int, int> > v;
    vector<queue<pair<int, int>>> q(11);
    for (int m, id = 1; id <= n; id++) {
        scanf("%d" , &m);
        for ( int x , y ; m; m--) {
            scanf(" T%d-%d" , &x ,&y );
            if (a[x] > 0) a[x]--, v.emplace_back(x, id, y);
            else q[x].emplace(id, y);
        }
        getchar();
    }
    for (int i = 1; i <= 10; i++) {
        if (a[i] == 0) continue;
        for (int j = i + 1; a[i] && j <= 10; j++) {
            while (!q[j].empty() && a[i])
                a[i]--, v.emplace_back(j, q[j].front().first, q[j].front().second), q[j].pop();
        }
        for (int j = i - 1; a[i] && j >= 1; j--) {
            while (!q[j].empty() && a[i])
                a[i]--, v.emplace_back(j, q[j].front().first, q[j].front().second), q[j].pop();
        }
    }
    sort(v.begin(), v.end());
    for( auto it : v )
        printf("%d-%d%c" , get<1>(it) , get<2>(it) , " \n"[it==v.back()]);
    return 0;
}

L2-3 简单的数数

比较典的 dp题，\(f[i][j]\)表示前\(i\)个数够成\(j\)的方案数

#include <bits/stdc++.h>

using namespace std;

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

const int mod = 998244353;

int main() {
    int n = read();
    vector<array<int, 10> > f(n + 1);;
    for (int i = 1, x; i <= n; i++) {
        x = read();
        if (i == 1) {
            f[i][x] = 1;
            continue;
        }
        for (int j = 0; j < 10; j++) {
            if (f[i - 1][j]) {
                f[i][(j + x) % 10] = (f[i][(j + x) % 10] + f[i - 1][j]) % mod;
                f[i][(j * x) % 10] = (f[i][(j * x) % 10] + f[i - 1][j]) % mod;
            }
        }
    }
    for (int i = 0; i <= 9; i++)
        cout << f[n][i] << "\n";
    return 0;
}

L2-4 回家日

首先，是我赛时的做法，就是按照解封的顺序把点加入到图中，如果这个点通过一条边直接相连的点也已经解封就把边也加入进来，然后用并查集动态的维护联通性，每次更新完一个点就把逐个判断剩余的点中是否有和\(k\)联通的点

#include <bits/stdc++.h>

using namespace std;

class dsu {
private:
    vector<int> fa;
public:
    dsu(int n = 1) {
        fa = vector<int>(n + 1, -1), fa[0] = 0;
    }

    int getfa(int x) {
        if (fa[x] < 0) return x;
        return fa[x] = getfa(fa[x]);
    }

    void merge(int x, int y) {
        x = getfa(x), y = getfa(y);
        if (x == y) return;
        if (fa[x] > fa[y]) swap(x, y);
        fa[x] += fa[y], fa[y] = x;
    }

    bool check(int x, int y) {
        x = getfa(x), y = getfa(y);
        return (x == y);
    }
};

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

int main() {
    int n = read(), m = read(), k = read();
    dsu d(n);
    vector<int> a(n + 1), p(n + 1, -1);
    for (int i = 1; i <= n; i++) a[i] = read();
    vector<vector<int>> e(n + 1, vector<int>());
    for (int u, v; m; m--)
        u = read(), v = read(), e[u].push_back(v), e[v].push_back(u);
    int t = 1;
    for (; t <= n; t++) {
        p[a[t]] = 0;
        for (auto v: e[a[t]]) {
            if (p[v] == 0) d.merge(v, a[t]);
        }
        if (a[t] == k) break;
    }
    unordered_set<int> st;
    for (int i = 1; i <= n; i++) {
        if (p[i] == 0 && d.check(i, k) == true) p[i] = t;
        else st.emplace(i);
    }
    t ++;
    for (; t <= n; t++){
        p[ a[t] ] = 0;
        for (auto v: e[a[t]]) {
            if (p[v] >= 0) d.merge(v, a[t]);
        }
        vector<int> cur;
        for( auto j : st )
            if( p[j] == 0 && d.check( j , k ) == true )
                p[j] = t , cur.push_back(j);
        for( auto j : cur )
            st.erase(j);
    }
    for( int i = 1 ; i <= n ; i ++ )
        printf("%d%c" , p[i] , (" \n"[i==n]));
    return 0;
}

然后正确实际上，我们可以用 bfs 的方式来维护这张图\(dis[i]\)表示\(i\)到\(k\)路径上的最大值，从起点开始搜索就好了。

#include <bits/stdc++.h>

using namespace std;

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

int32_t main() {
    int n = read(), m = read(), sta = read();
    vector<int> a(n + 1);

    for (int i = 1 , x ; i <= n; i++) x = read() , a[x] = i ;
    vector<vector<int>> e(n + 1);
    for (int u, v; m; m--)
        u = read(), v = read(), e[u].push_back(v), e[v].push_back(u);
    vector<int>dis(n + 1, INT_MAX);
    dis[sta] = a[sta];
    queue<int> q;
    q.emplace(sta);
    while (!q.empty()) {
        auto u = q.front(); q.pop();
        for (auto v: e[u]) {
            if( dis[v] <= max( a[v] , dis[u] ) ) continue;
            dis[v] = max( a[v] , dis[u] ) , q.emplace( v );
        }
    }
    for( int i = 1 ; i <= n ; i ++ )
        printf("%d%c" , dis[i] , " \n"[i==n] );
    return 0;
}

然后可以优化这个搜索过程，通过优先队列优化搜索顺序，并且保证每个点只扩展一次。

#include <bits/stdc++.h>

using namespace std;

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

int32_t main() {
    int n = read(), m = read(), sta = read();
    vector<int> a(n + 1);

    for (int i = 1 , x ; i <= n; i++) x = read() , a[x] = i ;
    vector<vector<int>> e(n + 1);
    for (int u, v; m; m--)
        u = read(), v = read(), e[u].push_back(v), e[v].push_back(u);
    vector<int>dis(n + 1, INT_MAX);
    dis[sta] = a[sta];
    queue<int> q;
    q.emplace(sta);
    while (!q.empty()) {
        auto u = q.front(); q.pop();
        for (auto v: e[u]) {
            if( dis[v] <= max( a[v] , dis[u] ) ) continue;
            dis[v] = max( a[v] , dis[u] ) , q.emplace( v );
        }
    }
    for( int i = 1 ; i <= n ; i ++ )
        printf("%d%c" , dis[i] , " \n"[i==n] );
    return 0;
}

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