題意:給你某個字串的\(n-1\)個字首和\(n-1\)個字尾,保證每個所給的字首字尾長度從\([1,n-1]\)都有,問你所給的子串是字首還是字尾.

題解:這題最關鍵的是那兩個長度為\(n-1\)的子串,我們只要判斷哪個是字首就行了,然後再遍歷一遍所給的子串,用長度為\(n-1\)的字首子串來判斷是子串是字首還是字尾.

程式碼:

int n;
string s[N];
bool vis[N];
int cnt;

int main() {
    ios::sync_with_stdio(false);cin.tie(0);
    cin>>n;

    string s1="";
    string s2="";
    string pre;
    for(int i=1;i<=2*n-2;++i){
        cin>>s[i];
        if(s1.size()<s[i].size()) s1=s[i];
        else if(s1.size()==s[i].size()) s2=s[i];
    }

    for(int i=1;i<=2*n-2;++i){
        if(s1.substr(0,s[i].size())==s[i]) cnt++;
    }

    if(cnt>=n-1 && s1.substr(1,s1.size())==s2.substr(0,s2.size()-1)) pre=s1;
    else pre=s2;

    for(int i=1;i<=2*n-2;++i){
        if(s[i]==pre.substr(0,s[i].size()) && !vis[s[i].size()]){
            printf("P");
            vis[s[i].size()]=true;
        }
        else printf("S");
    }

    return 0;
}