题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4622
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.
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：查询区间【l，r】的字符串有多少种不同的字串。
解法：这题也可以用后缀自动机解，这里只简单讲一下hash的做法。
首先我们可以枚举区间的长度L，然后枚举区间的起点l。然后我们可以考虑去重的情况，举个栗子吧：比如某字符串为abcabc，我们可以发现abc在区间【1,6】中出现了两次，所以我们应该减掉一个。去重的话，可以利用hash判断当前字符串之前是否出现过，那么问题来了：如果我们定义一个map<ull,int>,来判断的话，可能会出现超时的情况（目前还不知道为啥，难道是使用map的原因吗），做到这里我们可以建一个图，图的边权值为此时字符串的hash值，具体的实现看代码中的注释吧。
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<stack>
#include<cstdio>
#include<map>
#include<set>
#include<string>
#include<queue>
using namespace std;
#define inf 0x3f3f3f3f
#define ri register int
typedef long long ll;
typedef unsigned long long ull;
inline ll gcd(ll i,ll j){
return j==0?i:gcd(j,i%j);
}
inline ll lcm(ll i,ll j){
return i/gcd(i,j)*j;
}
inline void output(int x){
if(x==0){putchar(48);return;}
int len=0,dg[20];
while(x>0){dg[++len]=x%10;x/=10;}
for(int i=len;i>=1;i--)putchar(dg[i]+48);
}
inline void read(int &x){
char ch=x=0;
int f=1;
while(!isdigit(ch)){
ch=getchar();
if(ch=='-'){
f=-1;
}
}
while(isdigit(ch))
x=x*10+ch-'0',ch=getchar();
x=x*f;
}
const ull p=131;
const int maxn=2e3+5;
ull bit[maxn];
char ch[maxn];ull ha[maxn];
const ull mod=1e5+7;
ull getha(int l,int r){
if(l==r)return ch[l];
return ha[r]-ha[l-1]*bit[r-l+1];
}
struct gra{
int head[mod+5];
int pos[maxn];//当前字符出现的最晚的位置
int next[maxn];
ull edg[maxn];//hash值
int num;
void init(){
num=0;
memset(head,0,sizeof(head));
}
int find(ull val,int id){
int u=val%mod;
for(int i=head[u];i!=0;i=next[i]){
if(edg[i]==val){
int tem=pos[i];
pos[i]=id;
return tem;
}
}
num++;
edg[num]=val;
pos[num]=id;
next[num]=head[u];
head[u]=num;
return 0;
}
}h;
int dp[maxn][maxn];
int main(){
int t,q,l,r;
scanf("%d",&t);
bit[0]=1;
for(int i=1;i<maxn;i++){
bit[i]=bit[i-1]*p;
}
while(t--){
scanf("%s",ch+1);
int len=strlen(ch+1);
memset(dp,0,sizeof(dp));
ha[0]=1;
for(int i=1;i<=len;i++){
ha[i]=ha[i-1]*p+ch[i];
}
for(int i=1;i<=len;i++){
h.init();
for(int lm=1;lm+i-1<=len;lm++){
ull tem=getha(lm,lm+i-1);
int pos=h.find(tem,lm);
dp[pos][lm+i-1]--;
dp[lm][lm+i-1]++;
}
}
for(int i=len;i>=1;i--){
for(int j=i+1;j<=len;j++){
dp[i][j]+=dp[i][j-1]+dp[i+1][j]-dp[i+1][j-1];
}
}
scanf("%d",&q);
while(q--){
scanf("%d%d",&l,&r);
printf("%d
",dp[l][r]);
}
}
return 0;
}
使用map判重：
建图判重：
（区别太大了吧QAQ）