poj3261 Milk Patterns （后缀数组）

给n（n<=20000）,k和一个长度为n的由0..1,000,000 组成的字符串，问这个串中最长的重复k次的子串的长度。依然是后缀数组的模板，看到每个字符范围挺大，还是离散化一下比较安全。用离散化后的串构造出height数组后，因为构造好后缀数组后，重复出现的子串所在的后缀串一定是相邻的，所以直接枚举起点i终点j，长度即重复的次数k，用RMQ求一个i..j的最小值，记录一下这些最小值中的最大值输出即可。 不过这道题好像用字符串哈希更快一点，倍增的sa跑了100ms+...下面贴一个sa的代码。

#include <iostream>#include <algorithm>#include <cmath>#include <cstdio>#include <algorithm>#include <stack>#include <queue>#include <map>#include <string>#include <cstring>#include <string>using namespace std;typedef long long ll;const int maxn=40000+40;int s[maxn],rs[maxn];int sa[maxn],t[maxn],t2[maxn],c[maxn];int n,m,k;char s1[maxn],s2[maxn];int rank[maxn],height[maxn];int l1,l2;void getheight(int n){    int i,j,k=0;    for (i=0; i<=n; i++) rank[sa[i]]=i;    for (i=0; i<n; i++)    {        if (k) k--;        int j=sa[rank[i]-1];        while(s[i+k]==s[j+k]) k++;        height[rank[i]]=k;    }}void build_ss(int m,int n){    n++;    int i,*x=t,*y=t2;    for (int i=0; i<m; i++) c[i]=0;    for (int i=0; i<n; i++) c[x[i]=s[i]]++;    for (int i=1; i<m; i++) c[i]+=c[i-1];    for (int i=n-1; i>=0; i--)      sa[--c[x[i]]]=i;    for (int k=1; k<=n; k<<=1)    {        int p=0;        for (i=n-k; i<n; i++) y[p++]=i;        for (i=0; i<n; i++) if (sa[i]>=k) y[p++]=sa[i]-k;        for (i=0; i<m; i++) c[i]=0;        for (i=0; i<n; i++) c[x[y[i]]]++;        for (i=1; i<m; i++) c[i]+=c[i-1];        for (i=n-1; i>=0; i--) sa[--c[x[y[i]]]] = y[i];        swap(x,y);        p=1;        x[sa[0]]=0;        for (i=1; i<n; i++)        x[sa[i]]=(y[sa[i-1]]==y[sa[i]] && y[sa[i-1]+k]==y[sa[i]+k])? p-1 : p++;        if (p>=n) break;        m=p;    }}int d[maxn][20];void RMQ_init(){    for (int i=1; i<=n; i++)    d[i][0]=height[i];    for (int j=1; (1<<j)<=n; j++)     for (int i=1; i+(1<<(j-1))<=n; i++)     d[i][j]=min(d[i][j-1],d[i+(1<<(j-1))][j-1]);}int RMQ(int L,int R){    int k=0;    while ((1<<(k+1))<=R-L+1) k++;    return min(d[L][k],d[R-(1<<k)+1][k]);}void print(int k){    for (int i=k; i<n; i++) printf("%d ",s[i]);    printf("\n");}struct node{    int dt,id,neu;}a[20020];bool cmpid(node p,node q){    return p.id<q.id;}bool cmpdt(node p,node q){    if (p.dt!=q.dt) return p.dt<q.dt;    return p.id<q.id;}int main(){//    freopen("in.txt","r",stdin);    while(~scanf("%d%d",&n,&k))    {        for (int i=0; i<n; i++)        scanf("%d",&a[i].dt),a[i].id=i;        sort(a,a+n,cmpdt);        int rank=1;        a[0].neu=1;        for (int i=1; i<n; i++)        {            if (a[i].dt!=a[i-1].dt) rank++;            a[i].neu=rank;        }        sort(a,a+n,cmpid);        for (int i=0;i<n; i++)        s[i]=a[i].neu;        s[n]=0;        build_ss(rank+1,n);        getheight(n);        int i=1,j=i+k-1;        int ans=0;        RMQ_init();        while(j<=n)        {            ans=max(ans,RMQ(i+1,j));            i++;            j++;        }        printf("%d\n",ans);    }    return 0;}