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
题意
求区间不相同子串个数。
思路
模板题目，有好多种解法：字符串哈希、后缀数组、后缀自动机…
AC 代码
字符串哈希
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<cmath>
#include<algorithm>
#include<iostream>
using namespace std;
const int HASH = 10007;
const int MAXN = 2100;
struct hashmap
{
int head[HASH],next[MAXN],size;
unsigned long long state[MAXN];
int f[MAXN];
void init()
{
size=0;
memset(head,-1,sizeof(head));
}
int insert(unsigned long long val,int _id)
{
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;
}
} h;
const int seed = 13331;
unsigned long long P[MAXN];
unsigned long long S[MAXN];
char str[MAXN];
int ans[MAXN][MAXN];
int main()
{
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);
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]++;
ans[l][i+L-1]--;
}
}
for(int i=n; i>=0; i--)
for(int j=i; j<=n; j++)
ans[i][j]+=ans[i+1][j]+ans[i][j-1]-ans[i+1][j-1];
int Q;
scanf("%d",&Q);
while(Q--)
{
int a,b;
scanf("%d%d",&a,&b);
printf("%d\n",ans[a][b]);
}
}
return 0;
}
后缀数组
#include<algorithm>
#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=2010;
typedef long long ll;
char txt[maxn];
int sa[maxn],T1[maxn],T2[maxn],ct[maxn],he[maxn],rk[maxn],n,m;
void getsa(char *st)
{
int i,k,p,*x=T1,*y=T2;
for(i=0; i<m; i++) ct[i]=0;
for(i=0; i<n; i++) ct[x[i]=st[i]]++;
for(i=1; i<m; i++) ct[i]+=ct[i-1];
for(i=n-1; i>=0; i--)
sa[--ct[x[i]]]=i;
for(k=1,p=1; p<n; k<<=1,m=p)
{
for(p=0,i=n-k; i<n; i++) y[p++]=i;
for(i=0; i<n; i++) if(sa[i]>=k) y[p++]=sa[i]-k;
for(i=0; i<m; i++) ct[i]=0;
for(i=0; i<n; i++) ct[x[y[i]]]++;
for(i=1; i<m; i++) ct[i]+=ct[i-1];
for(i=n-1; i>=0; i--) sa[--ct[x[y[i]]]]=y[i];
for(swap(x,y),p=1,x[sa[0]]=0,i=1; i<n; i++)
x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+k]==y[sa[i]+k]?p-1:p++;
}
}
void gethe(char *st)
{
int i,j,k=0;
for(i=0; i<n; i++) rk[sa[i]]=i;
for(i=0; i<n-1; i++)
{
if(k) k--;
j=sa[rk[i]-1];
while(st[i+k]==st[j+k]) k++;
he[rk[i]]=k;
}
}
int main()
{
int t,i,q,le,ri,lcp,ml,ans,tp;
scanf("%d",&t);
while(t--)
{
scanf("%s",txt);
n=strlen(txt)+1;
m=150;
getsa(txt);
gethe(txt);
scanf("%d",&q);
while(q--)
{
scanf("%d%d",&le,&ri);
le--,ri--;
ans=lcp=ml=0;
for(i=1; i<n; i++)
{
lcp=min(lcp,he[i]);
ml=min(ml,lcp);
if(sa[i]>=le&&sa[i]<=ri)
{
ans+=ri-sa[i]+1-min(ri-sa[i]+1,ml);
lcp=he[i+1];
tp=min(lcp,ri-sa[i]+1);
ml=max(ml,tp);
}
}
printf("%d\n",ans);
}
}
return 0;
}
后缀自动机
#include<algorithm>
#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=2010<<1;
typedef long long ll;
struct node
{
node *f,*ch[30];
int len,way;
void init()
{
way=0;
memset(ch,0,sizeof ch);
}
}*root,*tail,sam[maxn<<1];
int tot,dp[2010][2010];
char txt[maxn];
void init()
{
tot=0;
root=tail=&sam[tot++];
root->f=NULL;
root->len=0;
root->init();
root->way=1;
}
void Insert(int c)
{
node *p=tail,*np=&sam[tot++];
np->init(),np->len=p->len+1;
while(p&&!p->ch[c])
{
np->way+=p->way;
p->ch[c]=np,p=p->f;
}
tail=np;
if(!p)
np->f=root;
else
{
if(p->ch[c]->len==p->len+1)
np->f=p->ch[c];
else
{
node *q=p->ch[c],*nq=&sam[tot++];
*nq=*q;
nq->len=p->len+1;
nq->way=0;
q->f=np->f=nq;
while(p&&p->ch[c]==q)
{
p->ch[c]->way-=p->way;
nq->way+=p->way;
p->ch[c]=nq,p=p->f;
}
}
}
}
int main()
{
int i,j,t,len,q,le,ri;
scanf("%d",&t);
while(t--)
{
scanf("%s",txt);
len=strlen(txt);
for(i=0; i<len; i++)
{
init();
for(j=i; j<len; j++)
{
Insert(txt[j]-'a');
dp[i][j]=tail->way;
}
for(j=i+1; j<len; j++)
dp[i][j]+=dp[i][j-1];
}
scanf("%d",&q);
while(q--)
{
scanf("%d%d",&le,&ri);
printf("%d\n",dp[le-1][ri-1]);
}
}
return 0;
}
