POJ 3261 Milk Patterns

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题意：给出一个字符串，问最长的可重叠的出现K次的子串的长度

子串重复，考虑用到后缀数组、LCP（最长公共前缀）。

二分答案，每一次按照子串排名依次遍历，看是否存在连续K个子串的height值均超过二分结果。

代码应该是挺简洁的→_→

#include<cstdio>#include<algorithm> #include<cstring>#define MAXN 20005#define MAX 1000005using namespace std;int n, k, s[MAXN];int sum[MAX], sa[MAXN], tsa[MAXN], rank[2*MAXN], trank[MAXN], height[MAXN];void fsort(int j){    memset(sum,0,sizeof(sum));    for(int i = 1; i <= n; i++)sum[rank[i+j]]++;    for(int i = 1; i < MAX; i++)sum[i]+=sum[i-1];    for(int i = n; i >= 1; i--)tsa[sum[rank[i+j]]--]=i;    memset(sum,0,sizeof(sum));    for(int i = 1; i <= n; i++)sum[rank[i]]++;    for(int i = 1; i < MAX; i++)sum[i]+=sum[i-1];    for(int i = n; i >= 1; i--)sa[sum[rank[tsa[i]]]--]=tsa[i];}void SA(){    memset(sum,0,sizeof(sum));    for(int i = 0; i <= n; i++)trank[i]=s[i];    for(int i = 1; i <= n; i++)sum[trank[i]]++;    for(int i = 1; i < MAX; i++)sum[i]+=sum[i-1];    for(int i = n; i >= 1; i--)sa[sum[trank[i]]--]=i;    rank[sa[1]]=1;    for(int i = 2, q = 1; i <= n; i++)    {        if(trank[sa[i]]!=trank[sa[i-1]])q++;        rank[sa[i]]=q;    }    for(int j = 1; j <= n; j *= 2)    {        fsort(j);        trank[sa[1]]=1;        for(int i = 2, q = 1; i <= n; i++)        {            if(rank[sa[i]]!=rank[sa[i-1]] || rank[sa[i]+j]!=rank[sa[i-1]+j])q++;            trank[sa[i]]=q;        }        for(int i = 1; i <= n; i++)            rank[i]=trank[i];    }}void LCP(){    int k = 0;    for(int i = 1; i <= n; i++)    {        if(k>0)k--;        int j = sa[rank[i]-1];        while(s[i+k]==s[j+k])k++;        height[rank[i]]=k;    }}int check(int x){    int cnt=1;    for(int i = 2; i <= n; i++)    {        if(height[i]>=x)            cnt++;        else            cnt=1;        if(cnt>=k)return true;    }    return false;}int main(){    scanf("%d%d",&n,&k);    for(int i = 1; i <= n; i++)        scanf("%d",&s[i]);    if(k==1)    {        sort(s+1,s+1+n);        int cnt=0;        for(int i = 1; i <= n; i++)        {            if(s[i]!=s[i-1])                cnt=0;            else cnt++;        }        printf("%d\n",cnt);        return 0;    }    s[0]=s[n+1]=MAX;    SA();    LCP();    int l = 1, r = n;    while(l<r)    {        int mid=(l+r+1)/2;        if(check(mid))l=mid;        else r=mid-1;    }    printf("%d",l);    return 0;}

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