题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3518
Problem Description
035 now faced a tough problem,his english teacher gives him a string,which consists with n lower case letter,he must figure out how many substrings appear at least twice,moreover,such apearances can not overlap each other.
Take aaaa as an example.”a” apears four times,”aa” apears two times without overlaping.however,aaa can’t apear more than one time without overlaping.since we can get “aaa” from [0-2](The position of string begins with 0) and [1-3]. But the interval [0-2] and
[1-3] overlaps each other.So “aaa” can not take into account.Therefore,the answer is 2(“a”,and “aa”).
Input
The input data consist with several test cases.The input ends with a line “#”.each test case contain a string consists with lower letter,the length n won’t exceed 1000(n <= 1000).
Output
For each test case output an integer ans,which represent the answer for the test case.you’d better use int64 to avoid unnecessary trouble.
Sample Input
aaaa ababcabb aaaaaa #
Sample Output
2 3 3
Source
2010 ACM-ICPC Multi-University
Training Contest（9）——Host by HNU
题意：
如题，求字符串中不重叠的重复出现至少两次的子串的个数！
PS:
运用后缀数组的性质，
先将后缀数组的sa、rank、height数组求出来！
然后在枚举重复子串的长度i，然后再在连续（至于为什么是连续的将在下面解释）且值大于所枚举的长度i的height数组中找出最大值和最小值，
如果最大值和最小值的差大于等于所枚举的长度i，那么答案增加一；
为什么需要连续的height数组呢？
因为这样可以不用再考虑重复计数的问题，如果有相同的前缀那么他们的rank排名必定是相邻的，那么height就是连续的，前缀不同的则再做另一次判断！
代码如下：
#include <cstdio>
#include <cstring>
const int N = 1017;
int wa[N], wb[N], wv[N], ws[N];
int rank[N]; //名次数组
int height[N]; //排名相邻的两个后缀的最长公共前缀
char str[N];
int s[N], sa[N]; //sa为后缀数组,n个后缀从小到大进行排序之后把排好序的后缀的开头位置
int Max(int a, int b)
{
return a > b ? a:b;
}
int Min(int a, int b)
{
return a < b ? a:b;
}
int cmp(int *r, int a, int b, int l)
{
return r[a]==r[b] && r[a+l]==r[b+l];
}
//get_sa函数的参数n代表字符串中字符的个数，这里的n里面是包括人为在字符串末尾添加的那个0的
//get_sa函数的参数m代表字符串中字符的取值范围，是基数排序的一个参数，
//如果原序列都是字母可以直接取128，
//如果原序列本身都是整数的话，则m可以取比最大的整数大1的值。
void get_sa(int *r, int *sa, int n, int m) //倍增算法
{
int i,j,p,*x=wa,*y=wb,*t;
for(i=0; i<m; i++) ws[i]=0;
for(i=0; i<n; i++) ws[x[i]=r[i]]++;
for(i=1; i<m; i++) ws[i]+=ws[i-1];
for(i=n-1; i>=0; i--) sa[--ws[x[i]]]=i; //对长度为1的字符串排序
for(p=1,j=1; p<n; j*=2,m=p)
{
for(p=0,i=n-j; i<n; i++) y[p++]=i;
for(i=0; i<n; i++) if(sa[i]>=j) y[p++]=sa[i]-j; //第二关键字排序结果
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]; //第一关键字排序
for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1; i<n; i++)
x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++; //更新rank数组
}
return;
}
void get_height(int *r, int *sa, int n) //求height数组
{
int i, j, k=0;
for(i=1; i<=n; i++) rank[sa[i]]=i;
for(i=0; i<n; height[rank[i++]]=k)
for(k?k--:0,j=sa[rank[i]-1]; r[i+k]==r[j+k]; k++);
return;
}
int main()
{
while(scanf("%s",str) != EOF)
{
int len = strlen(str);
if(str[0] == '#')
break;
for(int i = 0; i < len; i++)
s[i] = str[i]-'a'+1;
s[len] = 0;//这个赋值为0是关键所在
get_sa(s,sa,len+1,27);
get_height(s,sa,len);
//for(int i=0;i<len+1;i++) printf("%d %d\n",i,sa[i]);
int ans = 0;
for(int i = 1; i <= (len+1)/2; i++)
{
//查一半就好了，长度大于（len+1）/2的子串不可能重复俩次啦
//长度为i的重复子串
int minn = N;
int maxn = -1;
for(int j = 1; j <= len; j++)
{
if(height[j] >= i)//连续
{
minn = Min(minn,Min(sa[j-1],sa[j]));
maxn = Max(maxn,Max(sa[j-1],sa[j]));
}
else//若height的值不是连续大于i的
{
if(minn+i <= maxn)
ans++;
minn = N, maxn = -1;
}
}
if(minn+i <= maxn)
ans++;
}
printf("%d\n",ans);
}
return 0;
}
最后再贴两个讲解后缀数组很详细的链接：
2、http://www.cnblogs.com/staginner/archive/2012/02/02/2335600.html