【HDU - 4622】Reincarnation 【字符串HASH+dp 】

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.

题意 ：给个字符串，q次询问，每次询问区间[l,r]中不相同的子字符串个数。

分析讲解链接

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int>pii;
#define first fi
#define second se
#define  LL long long
#define ULL unsigned long long
#define fread() freopen("in.txt","r",stdin)
#define fwrite() freopen("out.txt","w",stdout)
#define CLOSE() ios_base::sync_with_stdio(false)

const int MAXN = 2000+11;
const int HASH = 10007;
const int MAXM = 1e6;
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;

struct HASHMAP{
int head[HASH],next[MAXN],size;
ULL state[MAXN];int f[MAXN];
void init(){
memset(head,-1,sizeof(head));
size=0;
}
int insert(ULL val,int id){// 返回上一个和当前字符串相同的左边界
int h=val%HASH;
for(int i=head[h];i!=-1;i=next[i]){
if(val==state[i]) {
int temp=f[i];
f[i]=id;
return temp;
}
}
f[size]=id; state[size]=val; next[size]=head[h];
head[h]=size++;
return 0; //当没有相同的字符串时候，返回0
}
fdab
}H;
const int SEED = 13331;
ULL P[MAXN],S[MAXN];
char str[MAXN];
int ans[MAXN][MAXN];

int main(){
CLOSE();
//  fread();
//  fwrite();
P[0]=1;
for(int i=1;i<MAXN;i++) P[i]=P[i-1]*SEED;
int T;scanf("%d",&T);
while(T--){
scanf("%s",str); int n=strlen(str);
S[0]=0;
for(int i=1;i<=n;i++)
S[i]=S[i-1]*SEED+str[i-1];

memset(ans,0,sizeof(ans));
for(int L=1;L<=n;L++){
H.init();
for(int i=1;i+L-1<=n;i++){
int l=H.insert(S[i+L-1]-S[i-1]*P[L],i);
ans[i][i+L-1]++;  //当前区间肯定加1
ans[l][i+L-1]--;//从上一个重复的位置开始到当前的右边界一定重复了，所以要减一。
}
}
for(int i=n;i>=1;i--)
for(int j=i+1;j<=n;j++)
ans[i][j]+=ans[i+1][j]+ans[i][j-1]-ans[i+1][j-1];// 当前的值是由后面的值确定的，所以要倒序来dp更新
int m,u,v;
scanf("%d",&m);
while(m--){
scanf("%d%d",&u,&v);
printf("%d\n",ans[u][v]);
}
}
return 0;
}