邝斌的ACM模板（后缀数组）

/*
*suffix array
*倍增算法  O(n*logn)
*待排序数组长度为n,放在0~n-1中，在最后面补一个0
*da(str ,n+1,sa,rank,height,  ,   );//注意是n+1;
*例如：
*n   = 8;
*num[]   = { 1, 1, 2, 1, 1, 1, 1, 2, $ };注意num最后一位为0，其他大于0
*rank[]  = { 4, 6, 8, 1, 2, 3, 5, 7, 0 };rank[0~n-1]为有效值，rank
必定为0无效 值
*sa[]    = { 8, 3, 4, 5, 0, 6, 1, 7, 2 };sa[1~n]为有效值，sa[0]必定为n是无效值
*height[]= { 0, 0, 3, 2, 3, 1, 2, 0, 1 };height[2~n]为有效值
*
*/
const int MAXN=20010;
int t1[MAXN],t2[MAXN],c[MAXN];//求SA数组需要的中间变量，不需要赋值
//待排序的字符串放在s数组中，从s[0]到s[n-1],长度为n,且最大值小于m,
//除s[n-1]外的所有s[i]都大于0，r[n-1]=0
//函数结束以后结果放在sa数组中
bool cmp(int *r,int a,int b,int l)
{
return r[a] == r[b] && r[a+l] == r[b+l];
}
void da(int str[],int sa[],int rank[],int height[],int n,int m)
{
n++;
int i, j, p, *x = t1, *y = t2;
//第一轮基数排序，如果s的最大值很大，可改为快速排序
for(i = 0; i < m; i++)c[i] = 0;
for(i = 0; i < n; i++)c[x[i] = str[i]]++;
for(i = 1; i < m; i++)c[i] += c[i-1];
for(i = n-1; i >= 0; i--)sa[--c[x[i]]] = i;
for(j = 1; j <= n; j <<= 1)
{
p = 0;
//直接利用sa数组排序第二关键字
for(i = n-j; i < n; i++)y[p++] = i;//后面的j个数第二关键字为空的最小
for(i = 0; i < n; i++)if(sa[i] >= j)y[p++] = sa[i] - j;
//这样数组y保存的就是按照第二关键字排序的结果
//基数排序第一关键字
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];
//根据sa和x数组计算新的x数组
swap(x,y);
p = 1;
x[sa[0]] = 0;
for(i = 1; i < n; i++)
x[sa[i]] = cmp(y,sa[i-1],sa[i],j)?p-1:p++;
if(p >= n)break;
m = p;
//下次基数排序的最大值
}
int k = 0;
n--;
for(i = 0; i <= n; i++)rank[sa[i]] = i;
for(i = 0; i < n; i++)
{
if(k)k--;
j = sa[rank[i]-1];
while(str[i+k] == str[j+k])k++;
height[rank[i]] = k;
}
}
int rank[MAXN],height[MAXN];
int RMQ[MAXN];
int mm[MAXN];
int best[20][MAXN];
void initRMQ(int n)
{
mm[0]=-1;
for(int i=1; i<=n; i++)         mm[i]=((i&(i-1))==0)?mm[i-1]+1:mm[i-1];
for(int i=1; i<=n; i++)best[0][i]=i;
for(int i=1; i<=mm
; i++)         for(int j=1; j+(1<<i)-1<=n; j++)
{
int a=best[i-1][j];
int b=best[i-1][j+(1<<(i-1))];
if(RMQ[a]<RMQ[b])best[i][j]=a;
else best[i][j]=b;
}
}
int askRMQ(int a,int b)
{
int t;
t=mm[b-a+1];
b-=(1<<t)-1;
a=best[t][a];
b=best[t][b];
return RMQ[a]<RMQ[b]?a:b;
}
int lcp(int a,int b)
{
a=rank[a];
b=rank[b];
if(a>b)swap(a,b);
return height[askRMQ(a+1,b)];
}
char str[MAXN];
int r[MAXN];
int sa[MAXN];
int main()
{
while(scanf("%s",str) == 1)
{
int len = strlen(str);
int n = 2*len + 1;
for(int i = 0; i < len; i++)r[i] = str[i];
for(int i = 0; i <
4000
; len; i++)r[len + 1 + i] = str[len - 1 - i];
r[len] = 1;
r
= 0;
da(r,sa,rank,height,n,128);
for(int i=1; i<=n; i++)RMQ[i]=height[i];
initRMQ(n);
int ans=0,st;
int tmp;
for(int i=0; i<len; i++)
{
tmp=lcp(i,n-i);//偶对称
if(2*tmp>ans)
{
ans=2*tmp;
st=i-tmp;
}
tmp=lcp(i,n-i-1);//奇数对称
if(2*tmp-1>ans)
{
ans=2*tmp-1;
st=i-tmp+1;
}
}
str[st+ans]=0;
printf("%s\n",str+st);
}
return 0;
}

/*
* 后缀数组
* DC3算法，复杂度O(n)
* 所有的相关数组都要开三倍
*/
const int MAXN = 2010;
#define F(x) ((x)/3+((x)%3==1?0:tb))
#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)
int wa[MAXN*3],wb[MAXN*3],wv[MAXN*3],wss[MAXN*3];
int c0(int *r,int a,int b)
{
return r[a] == r[b] && r[a+1] == r[b+1] && r[a+2] == r[b+2];
}
int c12(int k,int *r,int a,int b)
{
if(k == 2)   return r[a] < r[b] || ( r[a] == r[b] && c12(1,r,a+1,b+1) );
else return r[a] < r[b] || ( r[a] == r[b] && wv[a+1] < wv[b+1] );
}
void sort(int *r,int *a,int *b,int n,int m)
{
int i;
for(i = 0; i < n; i++)wv[i] = r[a[i]];
for(i = 0; i < m; i++)wss[i] = 0;
for(i = 0; i < n; i++)wss[wv[i]]++;
for(i = 1; i < m; i++)wss[i] += wss[i-1];
for(i = n-1; i >= 0; i--)
b[--wss[wv[i]]] = a[i];
}
void dc3(int *r,int *sa,int n,int m)
{
int i, j, *rn = r + n;
int *san = sa + n, ta = 0, tb = (n+1)/3, tbc = 0, p;
r
= r[n+1] = 0;
for(i = 0; i < n; i++)if(i %3 != 0)wa[tbc++] = i;
sort(r + 2, wa, wb, tbc, m);
sort(r + 1, wb, wa, tbc, m);
sort(r, wa, wb, tbc, m);
for(p = 1, rn[F(wb[0])] = 0, i = 1; i < tbc; i++)
rn[F(wb[i])] = c0(r, wb[i-1], wb[i]) ? p - 1 : p++;
if(p < tbc)dc3(rn,san,tbc,p);
else for(i = 0; i < tbc; i++)san[rn[i]] = i;
for(i = 0; i < tbc; i++) if(san[i] < tb)wb[ta++] = san[i] * 3;
if(n % 3 == 1)wb[ta++] = n - 1;
sort(r, wb, wa, ta, m);
for(i = 0; i < tbc; i++)wv[wb[i] = G(san[i])] = i;
for(i = 0, j = 0, p = 0; i < ta && j < tbc; p++)
sa[p] = c12(wb[j] % 3, r, wa[i], wb[j]) ? wa[i++] : wb[j++];
for(; i < ta; p++)sa[p] = wa[i++];
for(; j < tbc; p++)sa[p] = wb[j++];
} //str和sa也要三倍
void da(int str[],int sa[],int rank[],int height[],int n,int m)
{
for(int i = n; i < n*3; i++)
str[i] = 0;
dc3(str, sa, n+1, m);
int i,j,k = 0;
for(i = 0; i <= n; i++)rank[sa[i]] = i;
for(i = 0; i < n; i++)
{
if(k) k--;
j = sa[rank[i]-1];
while(str[i+k] == str[j+k]) k++;
height[rank[i]] = k;
}
}