Submission #675696

#	Time	Username	Problem	Language	Result	Execution time	Memory
675696		MrBrionix	Palindromic Partitions (CEOI17_palindromic)	C++17	100 / 100	5548 ms	98876 KiB

palindromic

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

constexpr int MAXN = 1 << 20, LOGN = 21;
int table[LOGN][MAXN];
void build(int N, vector<int> &V) {
    copy(V.begin(), V.begin()+N, table[0]);
    for (int j = 1; j < LOGN; j++) {
        for (int i = 0; i + (1 << j) <= N; i++) {
            table[j][i]=min(table[j-1][i],
                table[j-1][i+(1<<j)/2]);
        }
    }
}
int query(int l, int r) {
    int k = 31 - __builtin_clz(r - l); // [l, r)
    return min(table[k][l], table[k][r-(1 << k)]);
}

template<class T> using V = vector<T>; 
using vi = V<int>;
#define sz(x) int((x).size())
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define each(a,x) for (auto& a: x)
#define all(x) begin(x), end(x)

struct SuffixArray { 
    string S; int N; vi sa, isa, lcp; //N=sz(S)+1=n+1
    void init(string _S) { //$\mathcal{O}(n\log n)$
        N = sz(S = _S)+1; genSa(); genLcp(); 
    } //tested with $t\leq 10,n \leq 10^6$ (5.41 sec)
    void genSa() { // sa has size sz(S)+1, sa[0]=sz(S)
        sa = isa = vi(N); sa[0] = N-1; iota(1+all(sa),0);
        sort(1+all(sa),[&](int a,int b){return S[a]<S[b];});
        FOR(i,1,N) { int a = sa[i-1], b = sa[i];
            isa[b] = i > 1 && S[a] == S[b] ? isa[a] : i; }
        for (int len = 1; len < N; len *= 2) { // currently
            // sorted by first len chars
            vi s(sa), is(isa), pos(N); iota(all(pos),0); 
            each(t,s){
                int T=t-len;if(T>=0)sa[pos[isa[T]]++]=T;
            }
            FOR(i,1,N) { 
                int a = sa[i-1], b = sa[i]; 
                // verify that nothing goes out of bounds
                isa[b] = is[a]==is[b]&&
                is[a+len]==is[b+len]?isa[a]:i;
            }
        }//sa[i]= indice (0-based) d'inizio i-esimo... 
    }    //... prefisso più piccolo
    //lcp[i]=lcp tra sa[i] e sa[i+1] ($0\leq i \leq n-1$)
    void genLcp(){
        lcp=vi(N-1); int h=0; //lcp[0]=lcp(sz(S),sa[1])=0
        F0R(b,N-1) { int a = sa[isa[b]-1];
            while (a+h < sz(S) && S[a+h] == S[b+h]) ++h;
            lcp[isa[b]-1] = h; if (h) h--; }
        build(N-1,lcp); // serve rmq (del booklet)
        //if we cut off first chars of two strings with...
    }//...lcp h then remaining portions still have lcp h-1 
    //lcp of suffixes starting at a,b
    int getLCP(int a, int b){
        if (a == b) return sz(S)-a; // a,b sono 0-based
        int l = isa[a], r = isa[b]; if (l > r) swap(l,r);
        return query(l,r); // serve rmq (del booklet)
    }
};

void solve(){
    string s;
    int n;
    cin>>s;
    n=s.size();
    SuffixArray tmp;
    tmp.init(s);
    assert(tmp.lcp[0]==0);
    
    int last=0,ans=0;
    for(int i=0;i<s.size()/2;i++){
        if(s[last]==s[s.size()-1-i] && tmp.getLCP(last,s.size()-1-last-(i-last))>=(i-last+1)){
            last=i+1;
            ans+=2;
        }
    }
    
    if(last!=s.size()/2 || s.size()%2==1)ans++;
    cout<<ans<<"\n";
}

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

Compilation message (stderr)

palindromic.cpp: In function 'void solve()':
palindromic.cpp:79:18: warning: comparison of integer expressions of different signedness: 'int' and 'std::__cxx11::basic_string<char>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
   79 |     for(int i=0;i<s.size()/2;i++){
      |                 ~^~~~~~~~~~~
palindromic.cpp:86:12: warning: comparison of integer expressions of different signedness: 'int' and 'std::__cxx11::basic_string<char>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
   86 |     if(last!=s.size()/2 || s.size()%2==1)ans++;
      |        ~~~~^~~~~~~~~~~~
palindromic.cpp:71:9: warning: variable 'n' set but not used [-Wunused-but-set-variable]
   71 |     int n;
      |         ^

• Subtask 1: 15 / 15
• Subtask 2: 20 / 20
• Subtask 3: 25 / 25
• Subtask 4: 40 / 40