题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=4622
Reincarnation
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 2096 Accepted Submission(s): 715
Problem Description
Now you are back,and have a task to do:
Given you a string s consist of lower-case English letters only,denote f(s) as the number of distinct sub-string of s.
And you have some query,each time you should calculate f(s[l...r]), s[l...r] means the sub-string of s start from l end at r.
Given you a string s consist of lower-case English letters only,denote f(s) as the number of distinct sub-string of s.
And you have some query,each time you should calculate f(s[l...r]), s[l...r] means the sub-string of s start from l end at r.
Input
The first line contains integer T(1<=T<=5), denote the number of the test cases.
For each test cases,the first line contains a string s(1 <= length of s <= 2000).
Denote the length of s by n.
The second line contains an integer Q(1 <= Q <= 10000),denote the number of queries.
Then Q lines follows,each lines contains two integer l, r(1 <= l <= r <= n), denote a query.
For each test cases,the first line contains a string s(1 <= length of s <= 2000).
Denote the length of s by n.
The second line contains an integer Q(1 <= Q <= 10000),denote the number of queries.
Then Q lines follows,each lines contains two integer l, r(1 <= l <= r <= n), denote a query.
Output
For each test cases,for each query,print the answer in one line.
Sample Input
2
bbaba
5
3 4
2 2
2 5
2 4
1 4
baaba
5
3 3
3 4
1 4
3 5
5 5
Sample Output
3
1
7
5
8
1
3
8
5
1HintI won't do anything against hash because I am nice.Of course this problem has a solution that don't rely on hash.
Source
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题目意思：
给一个字符串，对每一个字符串，有非常多询问，询问给定区间不同子串的个数。
解题思路：
kmp+dp
普通的dp转移肯定超时。
sa[i][j]：表示以第j个字符開始可以往前最多的字符个数（如果为s个），要求满足【j-s+1,j】在【i,j-1】字符串区间出现。
这样要统计j開始的往前的情况，能够把字符串倒过来,把j作为第一个，然后1作为最后一个，求一遍next.然后更新sa[i][j] (i<j)
求出sa[i][j]后，就能够直接转移dp[i][j]=dp[i][j-1]+i-j+1-sa[i][j] //把第j个字符加上后，对整个子串个数的影响。减去在前面已经出现的。
代码：
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std; #define Maxn 2200 char sa1[Maxn];
char sa2[Maxn];
int next[Maxn];
int n,nn;
int sa[Maxn][Maxn],dp[Maxn][Maxn]; void getnext()
{
int j=0;
next[1]=0; for(int i=2;i<=nn;i++)
{
while(j>0&&sa1[j+1]-sa1[i])
j=next[j];
if(sa1[j+1]==sa1[i])
j++;
next[i]=j;
}
return ;
}
int main()
{
int t; scanf("%d",&t);
while(t--)
{
scanf("%s",sa1+1);
n=strlen(sa1+1);
for(int i=1;i<=n;i++)
sa2[i]=sa1[n+1-i]; for(int i=1;i<=n;i++)
{ for(int j=i;j<=n;j++)
sa1[j-i+1]=sa2[j];
nn=n-i+1;
getnext();
sa[n+1-i-1][n+1-i]=next[2]; for(int j=i+2;j<=n;j++)
sa[n+1-i-(j-i)][n+1-i]=max(next[j-i+1],sa[n+1-i-(j-i)+1][n+1-i]);
} for(int i=1;i<=n;i++)
{
dp[i][i]=1;
for(int j=i+1;j<=n;j++)
dp[i][j]=dp[i][j-1]+(j-i+1)-sa[i][j]; }
int q; scanf("%d",&q);
while(q--)
{
int a,b;
scanf("%d%d",&a,&b);
printf("%d\n",dp[a][b]);
}
}
return 0;
}