poj 3261 Milk Patterns

题目链接：http://poj.org/problem?id=3261

题目思路：后缀数组，求可重叠的重复k次子串。

#include<stdio.h>#include<stdlib.h>#include<string.h>#include<string>#include<queue>#include<algorithm>#include<vector>#include<stack>#include<list>//#include<iostream>#include<map>using namespace std;#define inf 0x3f3f3f3f#define M 1100000int max(int a,int b){return a>b?a:b;}int min(int a,int b){return a<b?a:b;}struct node{    int v,id;    bool operator<(const node a)const    {        return v<a.v;    }}a[M];int r[M],rank[M],sa[M],height[M];int ws[M],wv[M],wa[M],wb[M];bool cmp(int *r,int a,int b,int l){    return r[a]==r[b]&&r[a+l]==r[b+l];}void da(int n,int m){    int i,j,p;    int *x=wa,*y=wb;    for(i=0;i<n;i++)    {        a[i].v=r[i];        a[i].id=i;    }    sort(a,a+n);    p=1;    x[a[0].id]=0;    for(i=1;i<n;i++)    {        if(a[i].v==a[i-1].v) x[a[i].id]=p-1;        else x[a[i].id]=p++;    }     m=p;    for(i=0;i<m;i++) ws[i]=0;    for(i=0;i<n;i++) ws[x[i]]++;    for(i=1;i<m;i++) ws[i]+=ws[i-1];    for(i=n-1;i>=0;i--) sa[--ws[x[i]]]=i;    for(j=1,p=1;p<n;j*=2,m=p)    {        p=0;        for(i=n-j;i<n;i++) y[p++]=i;        for(i=0;i<n;i++) if(sa[i]>=j) y[p++]=sa[i]-j;        for(i=0;i<n;i++) wv[i]=x[y[i]];        for(i=0;i<m;i++) ws[i]=0;        for(i=0;i<n;i++) ws[wv[i]]++;        for(i=1;i<m;i++) ws[i]+=ws[i-1];        for(i=n-1;i>=0;i--) sa[--ws[wv[i]]]=y[i];        p=1;        swap(x,y);        x[sa[0]]=0;        for(i=1;i<n;i++)        {            if(cmp(y,sa[i],sa[i-1],j)) x[sa[i]]=p-1;            else x[sa[i]]=p++;        }    }}void calheight(int n){    int i,j,k;    for(i=1;i<=n;i++) rank[sa[i]]=i;    k=0;    for(i=0;i<n;i++)    {        int tmp=sa[rank[i]-1];        for(j=k;r[i+k]==r[tmp+k];k++);        height[rank[i]]=k;        k?--k:0;    }}int check(int n,int len,int k){    int num=0,i;    for(i=2;i<=n;i++)    {        if(height[i]<len)            num=0;        else        {            num++;            if(num>=k-1) return 1;        }    }    return 0;}int main(){    int i,n,k,left,right,mid;    while(scanf("%d%d",&n,&k)!=EOF)    {        for(i=0;i<n;i++) scanf("%d",&r[i]);        r[n]=0;        da(n+1,1000100);        calheight(n);        left=2;right=n;        while(left<=right)        {            mid=(left+right)>>1;            if(check(n,mid,k)) left=mid+1;            else right=mid-1;        }        printf("%d\n",right);    }}