2022 Shanghai Collegiate Programming Contest

1|0A. Another A+B Problem

暴力枚举出所有的情况，然后特判一下

#include <bits/stdc++.h>

using namespace std;

set<string> res;
int t;
string a , b , c = "????????";

#define f(d) ((d[0]-'0')*10 + d[1]-'0' + (d[3]-'0')*10 + d[4]-'0' == (d[6]-'0')*10 + d[7]-'0' )

void dfs(string p, const vector<int> &flag, int i) {
    if (p.size() > t) return;
    if (p.size() == t) {
        sort(p.begin(), p.end());
        do {
            string d = c;
            for (int i = 0, j = 0; i < 8; i++)
                if (d[i] == '?') d[i] = p[j], j++;
            int ss = 0;
            for( int i = 0 ; i < 8 ; i ++ )
                if( d[i] == a[i] && (b[i] == 'P' || b[i] == 'B') ){
                    ss = 1;
                    break;
                }
            if( ss ) continue;
            if (f(d))res.insert(d);
        } while (next_permutation(p.begin(), p.end()));
        return;
    }

    if (i >= 10) return;

    if (flag[i]) {
        dfs(p, flag, i + 1);
        return;
    }

    while (p.size() <= t) {
        dfs(p, flag, i + 1);
        p.push_back(i + '0');
    }
    return;
}

int main() {
    cin >> a >> b;
    vector<int> cnt(10);
    vector<int> flag(10, 0);
    for (int i = 0; i < 8; i++) {
        if (b[i] == 'G') c[i] = a[i];
        else if (b[i] == 'B') flag[a[i] - '0'] = 1;
        else cnt[a[i] - '0']++;
    }

    for (auto i: c) if (i == '?') t++;
    string p = "";
    for (int i = 0; i <= 9; i++) {
        for (; cnt[i]; cnt[i]--) p.push_back(i + '0');
    }
    dfs(p, flag, 0);

    cout << res.size() << "\n";
    for( auto i : res )
        cout << i << "\n";
    return 0;
}

2|0E. Expenditure Reduction

预处理出每个字母的位置，然后 dp 转移一下

#include <bits/stdc++.h>

using namespace std;
#define int long long

int dp[305];
void solve(){
    string a,b;
    cin>>a>>b;
    int n=a.size();
    int m=b.size();
    a=' '+a;
    b=' '+b;
    vector<int>mp[300];
    for(int i=m;i>=1;i--){
        mp[b[i]].push_back(i);
    }
    for(int i=1;i<=m;i++)dp[i]=-0x3f3f3f3f;
    int ans=0x3f3f3f3f3f,l=0;
    for(int i=1;i<=n;i++){
        dp[0]=i;
        for(auto it:mp[a[i]]){
            dp[it]=max(dp[it],dp[it-1]);
        }
        if(i-dp[m]+1<=ans){
            ans=i-dp[m]+1;
            l=dp[m];
        }
    }
    for(int i=0;i<ans;i++){
        cout<<a[l+i];
    }
    cout<<"\n";
    //cout<<ans<<"\n";
}
int32_t main() {
    int T;
    cin>>T;
    while(T--){
        solve();
    }
    return 0;
}

3|0G. Gua!

计算出可能造成伤害的最大值。

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

void solve(){
    double b,r,d,s;
    cin>>b>>r>>d>>s;
    if(d==0){
        cout<<"ok\n";
        return ;
    }
    if(r==0||b==0) {
        cout << "gua!\n";
        return ;
    }
    if(b+b*(long long)(s*r/60)>=d)cout<<"ok"<<endl;
    else cout<<"gua!"<<endl;

}

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

4|0H. Heirloom Painting

统计出所有颜色相同的段，长度除以 $k$ 上取证就是这一段需要的染色次数。注意最后一次染色的段的长度一定大于等于 $k$ 。

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

void solve(){
    int n , m , k;
    cin >> n >> m >> k;
    vector<int> a(n);
    for( auto & i : a ) cin >> i;
    vector<pair<int,int>> cnt;
    int t = 1 , last = a[0];
    for( int i = 1 ; i < n ; i ++ ){
        if( a[i] == last ) t ++;
        else {
            cnt.emplace_back( last , t );
            t = 1 , last = a[i];
        }
    }
    cnt.emplace_back( last , t );
    if( cnt.size()>=2&&cnt.back().first == cnt.front().first ){
        cnt.front().second += cnt.back().second;
        cnt.pop_back();
    }
    int f = 0 , res = 0;

    for( auto [l,t] : cnt ){
        if( t >= k ) f = 1;
        res += (t+k-1) / k;
    }
    if( f ) cout << res << "\n";
    else cout << "-1\n";
}

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

5|0M. My University Is Better Than Yours

这题应该是图论比较板的题？但是我图论会的不太多，所以学了一个不用图论的做法。首先我们每次遍历一列。把这列中所有没有访问过的点加入到队列中。如果累计访问过的点的数量等于当前的列数，说明当前队列中的点构成了一个环，更新队列中点的答案后，把队列清空即可。

#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;
    cin >> n >> m;
    vector<vector<int>> a( m+1 , vector<int>(n+1) );
    for( int i = 1 ; i <= m ; i ++ )
        for( int j = 1 ; j <= n ; j ++ )
            cin >> a[i][j];

    vector<int> q , vis( n+1  ) , res( n+1 );
    for( int i = 1 , cnt = 0 , rank = 1 ; i <= n ; i ++ ){
        for( int j = 1 ; j <= m ; j ++ )
            if( !vis[ a[j][i] ] ) vis[a[j][i]] = 1 , cnt ++ , q.push_back( a[j][i] );
        if( cnt == i ){
            for( auto j : q )
                res[j] = n - rank;
            q = vector<int>() , rank = i + 1;
        }
    }
    for( int i = 1 ; i <= n ; i ++ )
        cout << res[i] << " ";
    return 0;
}

6|0N. Nine Is Greater Than Ten

#include <bits/stdc++.h>

using namespace std;
#define int long long
const int maxn=1e6+5;


void solve(){
    string a,b;
    cin>>a>>b;
    if(a>b){
        cout<<a<<">"<<b;
    }
    else if(a==b)cout<<a<<"="<<b;
    else cout<<a<<"<"<<b;
}

int32_t main(){
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int T=1;
    while(T --)
        solve();
    return 0;
}