HDU 2328 Corporate Identity（后缀数组）
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2328
题意：给出n个串，求n个串的最长的字典序最小的公共子串？
思路：二分。
int r[N],sa[N],wa[N],wb[N],wd[N],rank[N],h[N];
int cmp(int *r,int a,int b,int len)
{
return r[a]==r[b]&&r[a+len]==r[b+len];
}
void da(int *r,int *sa,int n,int m)
{
int i,j,p,*x=wa,*y=wb,*t;
FOR0(i,m) wd[i]=0;
FOR0(i,n) wd[x[i]=r[i]]++;
FOR1(i,m-1) wd[i]+=wd[i-1];
FORL0(i,n-1) sa[--wd[x[i]]]=i;
for(j=1,p=1;p<n;j<<=1,m=p)
{
p=0;
FOR(i,n-j,n-1) y[p++]=i;
FOR0(i,n) if(sa[i]>=j) y[p++]=sa[i]-j;
FOR0(i,m) wd[i]=0;
FOR0(i,n) wd[x[i]]++;
FOR1(i,m-1) wd[i]+=wd[i-1];
FORL0(i,n-1) sa[--wd[x[y[i]]]]=y[i];
t=x;x=y;y=t;p=1;x[sa[0]]=0;
FOR1(i,n-1) x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
}
}
void calHeight(int *r,int *sa,int n)
{
int i,j,k=0;
FOR1(i,n) rank[sa[i]]=i;
FOR0(i,n)
{
if(k) k--;
j=sa[rank[i]-1];
while(i+k<n&&j+k<n&&r[i+k]==r[j+k]) k++;
h[rank[i]]=k;
}
}
int n,m,b[N];
char s[N];
int start;
int OK(int x)
{
int i,j,k,L,R,hash[4005];
FOR1(i,n)
{
L=i;
while(L<=n&&h[L]<x) L++;
if(L>n) break;
R=L;
while(R<=n&&h[R]>=x) R++;
clr(hash,0);
FOR(j,L-1,R-1) if(b[sa[j]]!=-1) hash[b[sa[j]]]=1;
FOR1(j,m) if(!hash[j]) break;
if(j>m)
{
start=sa[L-1];
return 1;
}
i=R;
}
return 0;
}
void print(int L)
{
int i;
FOR(i,start,start+L-1) putchar(r[i]);
puts("");
}
void deal()
{
int low=1,high=n,mid;
while(low<=high)
{
mid=(low+high)>>1;
if(OK(mid)) low=mid+1;
else high=mid-1;
}
if(low<=n&&OK(low)) print(low);
else if(high>=1&&OK(high)) print(high);
else puts("IDENTITY LOST");
}
int main()
{
while(scanf("%d",&m),m)
{
int i,k,j,L=0,x=130;
FOR1(j,m)
{
RD(s); k=strlen(s);
FOR0(i,k) r[L++]=s[i],b[L-1]=j;
b[L]=-1;
r[L++]=x++;
}
r[L]=0;
da(r,sa,L+1,x);
calHeight(r,sa,L);
n=L;
deal();
}
return 0;
}