Luo Gu P2463 [SDOI2008] Sandy card (suffix array SA + differential + half of the answer)

Topic links: https://www.luogu.org/problem/P2463

[Title] Italy

Obtaining N identical sub-strings appear in both the length of the longest string, the substring same as defined in the title: all elements plus a number into another, then the two identical strings can be very simply obtained, the difference i.e., the same as the same string.

[Thinking]

Of course, is to the N strings are differential, then the combined string into a string of N, can be connected end to end, but to mark those which belong to a number of Mi (behind To Check), it is to be noted here, remember separates the string,

Or will WA, where I is the partition between the string plus 0, the merger is completed, the subject can find the longest becomes the longest length of the substring is repeated N times do not overlap. Because if the string length k appears N times, then his prefix

Certainly there have been N times, so this question is monotone, can be half the answer.

[Check] Methods

Found continuous period of rank [l, r] are satisfied height [i]> = mid (l <i <= r) of the continuous section, the interval [l, r] are processed, it is determined that all the sa [i] (l <= i <= r) belongs Mi, if there are N different Mi condition is met, returns true,

Conversely returns false.

The Code

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

const int maxn = 1e6 + 5;
int n, t, M, b[maxn], num[110], a[maxn];
int sa[maxn], x[maxn], c[maxn], y[maxn], rk[maxn], height[maxn];
bool vis[maxn];
inline void get_sa(){
    int m = 2000;
    for( int i=1; i<=n; i++ ) ++c[x[i]=a[i]];
    for( int i=1; i<=m; i++ ) c[i] += c[i-1];
    for( int i=n; i; i-- ) sa[c[x[i]]--] = i;
    for( int k=1; k<=n; k<<=1 ){
        int now = 0;
        for( int i=n-k+1; i<=n; i++ ) y[++now] = i;
        for( int i=1; i<=n; i++ ) if(sa[i]>k) y[++now] = sa[i]-k;
        for( int i=0; i<=m; i++ ) c[i] = 0;
        for( int i=1; i<=n; i++ ) ++c[x[i]];
        for( int i=1; i<=m; i++ ) c[i] += c[i-1];
        for( int i=n; i; i-- ) sa[c[x[y[i]]]--] = y[i], y[i]=0;
        swap(x, y);
        x[sa[1]] = now = 1;
        for( int i=2; i<=n; i++ )
            x[sa[i]] = (y[sa[i]]==y[sa[i-1]] && y[sa[i]+k]==y[sa[i-1]+k]) ? now : ++now;
        if( now>=n ) return ;
        m = now;
    }
}

inline void get_height(){
    int k = 0;
    for( int i=1; i<=n; i++ ) rk[sa[i]] = i;
    for( int i=1; i<=n; i++ ){
        if( rk[i]==1 ) continue;
        if( k ) k--;
        int j = sa[rk[i]-1];
        while( j+k<=n && i+k<=n && a[j+k]==a[i+k] ) k++;
        height[rk[i]] = k;
    }
}

inline bool check1( int l, int r ){
    if( r-l+1<t ) return 0;                 //!!!!!!!r-l+1<t  不是r-l+1<n
    int tmp = 0;
    memset( vis, 0, sizeof(vis) );
    for( int i=l; i<=r; i++ )
        if( !vis[b[sa[i]]] ){
            tmp ++;
            vis[b[sa[i]]] = 1;
        }
    return tmp == t;
}

inline bool check( int x ){
    int l=1, r=1;
    while( l<=n ){
        while( height[r+1]>=x ) r++;
        if( check1(l, r) ) return 1;
        l = r+1; r = l;
    }
    return 0;
}

int main(){
    freopen("in.txt", "r", stdin);
    scanf("%d", &t);
    int len = 0;
    for( int i=1; i<=t; i++ ){
        scanf("%d", &M);
        for( int j=0; j<M; j++ ) scanf("%d", &num[j]);
        for( int j=1; j<M; j++ ){
            a[++n] = num[j]-num[j-1];           //进行差分
            b[n] = i;
        }
        a[++n] = 0;         //分隔不同的串
        b[n] = i;
    }
    // for( int i=1; i<=n; i++ ) cout << a[i] << endl;
    get_sa();
    get_height();
    // for( int i=1; i<=n; i++ ) cout << height[i] <<endl;
    int l=1, r=n;
    int ans=0;
    while( l<=r ){
        int mid = l+r>>1;
        if( check(mid) ) ans = mid, l=mid+1;
        else r = mid-1;
    }
    printf("%d\n", ans+1);

     return 0;
}