spoj 220 Relevant Phrases of Annihilation(n个串的最长公共重复2次子串)
地址：http://www.spoj.com/problems/PHRASES/
题意：给你n个串，求最长的公共重复子串
分析：跟两个串的最长的公共重复子串差不多，也是将所有串合成一个串，构造后缀数组，然后二分子串长度，在满足条件的区间里，求出n的串的下标的上下界的差值大于等于当前长度，所有串都满足说明存在解，否则无解。。。
代码：
/** head files*/
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>
using namespace std;
/** some operate*/
#define PB push_back
#define MP make_pair
#define REP(i,n) for(i=0;i<(n);++i)
#define UPTO(i,l,h) for(i=(l);i<=(h);++i)
#define DOWN(i,h,l) for(i=(h);i>=(l);--i)
#define MSET(arr,val) memset(arr,val,sizeof(arr))
#define MAX3(a,b,c) max(a,max(b,c))
#define MAX4(a,b,c,d) max(max(a,b),max(c,d))
#define MIN3(a,b,c) min(a,min(b,c))
#define MIN4(a,b,c,d) min(min(a,b),min(c,d))
/** some const*/
#define N 222222
#define M 222222
#define PI acos(-1.0)
#define oo 1111111111
/** some alias*/
typedef long long ll;
/** Global variables*/
/** some template names, just push ctrl+j to get it in*/
//manacher 求最长回文子串
//pqueue 优先队列
//combk n元素序列的第m小的组合和
//pmatrix n个点的最大子矩阵
//suffixarray 后缀数组
template <typename T, int LEN>
struct suffixarray
{
int str[LEN*3],sa[LEN*3];
int rank[LEN],height[LEN];
int id[LEN];
int len;
bool equal(int *str, int a, int b)
{
return str[a]==str[b]&&str[a+1]==str[b+1]&&str[a+2]==str[b+2];
}
bool cmp3(int *str, int *nstr, int a, int b)
{
if(str[a]!=str[b])return str[a]<str[b];
if(str[a+1]!=str[b+1])return str[a+1]<str[b+1];
return nstr[a+b%3]<nstr[b+b%3];
}
void radixsort(int *str, int *sa, int *res, int n, int m)
{
int i;
REP(i,m)id[i]=0;
REP(i,n)++id[str[sa[i]]];
REP(i,m)id[i+1]+=id[i];
DOWN(i,n-1,0)res[--id[str[sa[i]]]]=sa[i];
}
void dc3(int *str, int *sa, int n, int m)
{
#define F(x) ((x)/3+((x)%3==1?0:one))
#define G(x) ((x)<one?(x)*3+1:((x)-one)*3+2)
int *nstr=str+n, *nsa=sa+n, *tmpa=rank, *tmpb=height;
int i,j,k,len=0,num=0,zero=0,one=(n+1)/3;
REP(i,n)if(i%3)tmpa[len++]=i;
str[n]=str[n+1]=0;
radixsort(str+2, tmpa, tmpb, len, m);
radixsort(str+1, tmpb, tmpa, len, m);
radixsort(str+0, tmpa, tmpb, len, m);
nstr[F(tmpb[0])]=num++;
UPTO(i,1,len-1)
nstr[F(tmpb[i])]=equal(str,tmpb[i-1],tmpb[i])?num-1:num++;
if(num<len)dc3(nstr,nsa,len,num);
else REP(i,len)nsa[nstr[i]]=i;
if(n%3==1)tmpa[zero++]=n-1;
REP(i,len)if(nsa[i]<one)tmpa[zero++]=nsa[i]*3;
radixsort(str, tmpa, tmpb, zero, m);
REP(i,len)tmpa[nsa[i]=G(nsa[i])]=i;
i=j=0;
REP(k,n)
if(j>=len||(i<zero&&cmp3(str,tmpa,tmpb[i],nsa[j])))sa[k]=tmpb[i++];
else sa[k]=nsa[j++];
}
void initSA(T *s, int n,int m)
{
int i,j,k=0;
str[len=n]=0;
REP(i,n)str[i]=s[i];
dc3(str,sa,n+1,m);
REP(i,n)sa[i]=sa[i+1];
REP(i,n)rank[sa[i]]=i;
REP(i,n)
{
if(k)--k;
if(rank[i])for(j=sa[rank[i]-1];str[i+k]==str[j+k];++k);
else k=0;
height[rank[i]]=k;
}
}
};
suffixarray<char, N> msa;
char s[N],tmp[N];
int p[N],low[22],high[22];
int n;
bool ok(int m)
{
int i=0,j,k;
while(i<msa.len)
{
while(i<msa.len&&msa.height[i]<m)++i;
j=i-1;
UPTO(k,1,n)low[k]=msa.len,high[k]=-1;
do
{
k=msa.sa[j++];
low[p[k]]=min(low[p[k]],k);
high[p[k]]=max(high[p[k]],k);
}while(j<msa.len&&msa.height[j]>=m);
UPTO(k,1,n)
if(high[k]-low[k]<m)break;
if(k>n)return 1;
i=j;
}
return 0;
}
int main()
{
int i,j,k,m,t,ans;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
m=0;
UPTO(i,1,n)
{
scanf("%s",tmp);
k=strlen(tmp);
REP(j,k)
s[m+j]=tmp[j],p[m+j]=i;
p[m+k]=0;
s[m+k]=i;
m=m+k+1;
}
m--;
msa.initSA(s,m,256);
ans=i=0,j=m;
while(i<=j)
{
m=(i+j)>>1;
if(ok(m))ans=m,i=m+1;
else j=m-1;
}
printf("%d\n",ans);
}
return 0;
}