POJ 3261 Milk Patterns（SA 求最长可重叠重复K次子串）

题目链接：Click here~~

题意：

给一个字符串（int 型），求至少出现 K 次的最长子串，这 K 个子串可以重叠。

解题思路：

如果明白了上一题，这一题就很好想了。

分组后，一个组里面的后缀个数就代表着它们的 lcp 共出现了多少次，故统计好每组个数即可。5 min AC。

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;const int N = 2e4 + 5;int sa[N],rank[N],rank2[N],height[N],cnt[N*50],*x,*y;/*    * a radix_sort which is based on the y[].    * how ? ahhhh, the last reverse for is the solution.    * and the adjacant value of sa[] might have the same rank.*/void radix_sort(int n,int sz){    memset(cnt,0,sizeof(int)*sz);    for(int i=0;i<n;i++)        cnt[ x[ y[i] ] ]++;    for(int i=1;i<sz;i++)        cnt[i] += cnt[i-1];    for(int i=n-1;i>=0;i--)        sa[ --cnt[ x[ y[i] ] ] ] = y[i];}/*    * sa[i] represents the ith suffix string is which one.    * rank[i] represents the suffix string [i,n]'s rank.    * sz is the max_rank of text in that time.    * x[] represents the true pointer of rank[] in that time and it may be not unique.    * y[] is the location of text[] which is sorted by 2nd key in that time before swap(x,y).*/void get_sa(int text[],int n,int sz=1000001){    x = rank, y = rank2;    for(int i=0;i<n;i++)        x[i] = text[i], y[i] = i;    radix_sort(n,sz);    for(int len=1;len<n;len<<=1)    {        int yid = 0;        for(int i=n-len;i<n;i++)            y[yid++] = i;        for(int i=0;i<n;i++)            if(sa[i] >= len)                y[yid++] = sa[i] - len;        radix_sort(n,sz);        swap(x,y);        x[ sa[0] ] = yid = 0;        for(int i=1;i<n;i++)        {            if(y[ sa[i-1] ]==y[ sa[i] ] && sa[i-1]+len<n && sa[i]+len<n && y[ sa[i-1]+len ]==y[ sa[i]+len ])                x[ sa[i] ] = yid;            else                x[ sa[i] ] = ++yid;        }        sz = yid + 1;        if(sz >= n)            break;    }    for(int i=0;i<n;i++)        rank[i] = x[i];}/*    * height[] represents the longest common prefix of suffix [i-1,n] and [i,n].    * height[ rank[i] ] >= height[ rank[i-1] ] - 1.    ..... let's call [k,n] is the suffix which rank[k] = rank[i-1] - 1,    ...=> [k+1,n] is a suffix which rank[k+1] < rank[i]    ..... and the lcp of [k+1,n] and [i,n] is height[ rank[i] ] - 1.    ..... still unknow ? height[ rank[i] ] is the max lcp of rank[k] and rank[i] which rank[k] < rank[i].*/void get_height(int text[],int n){    int k = 0;    for(int i=0;i<n;i++)    {        if(rank[i] == 0)            continue;        k = max(0,k-1);        int j = sa[ rank[i]-1 ];        while(i+k<n && j+k<n && text[i+k]==text[j+k])            k++;        height[ rank[i] ] = k;    }}bool can(int k,int n,int cnt){    for(int i=0,j;i<n;i=j)    {        j = i + 1;        while(j < n && height[j] >= k)            j++;        if(j - i >= cnt)            return true;    }    return false;}int str[N];int main(){    //freopen("in.ads","r",stdin);    //freopen("out.ads","w",stdout);    int n,k;    while(~scanf("%d%d",&n,&k))    {        for(int i=0;i<n;i++)            scanf("%d",&str[i]);        get_sa(str,n);        get_height(str,n);        int l = 0 , r = n + 1;        while(l < r)        {            int mid = l+r >> 1;            if(can(mid,n,k))                l = mid + 1;            else                r = mid;        }        printf("%d\n",r-1);    }    return 0;}