[kmp+dp] hdu 4628 Pieces

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.

 

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.

 

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
1

Hint

      I won't do anything against hash because I am nice.Of course this problem has a solution that don't rely on hash.

 

Source

2013 Multi-University Training Contest 3

 

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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;
}