# Written before↵
↵
Border Theory,which was not well-known enough,should have been publicized together with KMP or similar algorithms.But nowadays lots of people don't know such a theory exist,making them unable to solve several tasks easily or even can't get a fast solution.So it's time to complete the blog as what I will do on this.↵
↵
# Ordinary Border Theory↵
↵
## Conclution↵
↵
For a string $S$, There are a squence of strings:↵
$$b_i(0 = b_1 < b_2 < b_3 < ... < b_n = |S|,S_{1,...,b_i} = S_{n - b_i + 1,...,n})$$↵
↵
it's garenteed that there exist a way to divide the sequence into $\operatorname{O}(\log |S|)$ continuous subsequence that each subsequence forms an arithmetic sequence.↵
↵
## Proof↵
↵
First of all,we found that if there's a string $T$ as a border of $S$ and $2|T| \ge |S|$, Then the sequence of where $T$ appears in $S$ forms an arthmetic sequence.↵
↵
<spoiler summary="proof">↵
If there's pairs of nearby places where $T$ matches $(i - p,i)$ and $(i,i + q)$ that $p \neq q$,we found that:↵
↵
- $p$ is the length of a border↵
- $q$ is the length of an another border↵
- The shorter border is a border of a larger border.↵
↵
Then we will found that if $p > q$, thenwe found pair $(i - q,i)$ exists,or $(i,i + p)$ exists while $p < q$, which are not valid when we deem pairs $(i - p,i)$,$(i,i + q)$ as nearby place that match,so $p = q$ garenteed.↵
</spoiler>↵
↵
Second,We found that all lengths of border $T$ that $2|T| \ge |S|$ will form an arthmetic sequence:↵
↵
<spoiler summary="proof">↵
if there's border $i + p,i,i - q$ as no border length $l$ that $i - q < l < i$ or $i < l < i + p$,we'll found:↵
↵
- $p,q$ are the lengths of borders.↵
- $i - p$ is the length of a border of the border with length $i$.↵
- $i$ is the length of a border of the border with length $i + q$.↵
↵
We will find $i - q < i - p < i$ or $i < i + q < i + p$, making it incorrect,so $p = q$↵
</spoiler>↵
↵
So there's a way that if $p < q,p,q \text{are the length of borders}$ not in the same continuous sequence, $2p < q$ exist,So the conclution is right.↵
↵
## Tasks↵
↵
[Link](https://www.luogu.com.cn/problem/P4156)↵
↵
Given at most $5$ cases in a test case,each cases gives a string $S(1 \le |S| \le 5 \times 10 ^ 5)$ and an integer $w(1 \le w \le 10 ^ {18})$, when we have a string $T$ we can make the new string by $|T| + |S|$ or if a suffix $A$ in $T$ is also the prefix of $S$, $T - A + S$ can be formed as there's a string of length $|T| - |A| + |S|$.Find the number of distinct length of strings in $[1,w]$ can be formed.↵
↵
<spoiler summary="solution">↵
In fact,if there's a way for an addition $x$,we can form a graph of $x$ nodes and deem other additions as edges to form a graph. We'll have to run a shortest paths algorithm to get the smallest lengths with garenteed modules.↵
↵
But the naive algorithm is too slow,as borders form $\operatorname{O}(\log |S|)$ distinct arthmetic sequences,we just start updating with an arthmetic sequence for $\operatorname{O}(\log |S|)$ rounds. Every time we start with the smallest node in each cycle and can update all the nodes in the cycle using a monotonous queue(as a sliding window problem) in $\operatorname{O}(x)$.↵
↵
At first we start with a graph with $|S|$ nodes and only node $0$ with a distance $|S|$, Then we may change $x$ to $x'$,it's also easy to start form the node with smallest distance in each cycle and update it in $\operatorname{O}(x + x')$.↵
↵
At last we get an $\operatorname{O}(|S|\log |S|)$ algorithm.↵
</spoiler>↵
↵
<spoiler summary="code">↵
↵
~~~~~↵
#include<bits/stdc++.h>↵
using namespace std;↵
int t,n,bor[500009],diff[500009],dt[500009],shlink[500009],q[500009],head,tail;↵
long long w,dp1[500009],dp2[500009];↵
int mx[500009];↵
const long long INF = 1998244353999911659ll;↵
char c[500009];↵
int gcd(int x,int y){↵
return y == 0 ? x : gcd(y,x % y);↵
}↵
int main(){↵
scanf("%d",&t);↵
while(t--){↵
scanf("%d %lld",&n,&w);↵
scanf(" %s",c + 1);↵
bor[1] = 0;↵
for(int i = 2; i <= n; i ++){↵
int t = bor[i - 1];↵
while(t != 0 && c[i] != c[t + 1])↵
t = bor[t];↵
if(c[i] == c[t + 1])↵
t ++;↵
bor[i] = t;↵
}↵
for(int i = 1; i <= n; i ++){↵
diff[i] = i - bor[i];↵
shlink[i] = bor[i];↵
if(diff[i] == diff[bor[i]])↵
shlink[i] = shlink[bor[i]];↵
}↵
//for(int i = n; i > 0; i = bor[i])↵
// printf("%d\n",bor[i]);↵
//puts("OK");↵
dp1[0] = n;↵
for(int i = 1; i < n; i ++){↵
dp1[i] = INF;↵
}↵
int lst = n;↵
for(int i = bor[n]; i != 0; i = shlink[i]){↵
int nw = n - i;↵
int ls = (i - shlink[i]) / diff[i] - 1;↵
for(int j = 0; j < nw; j ++){↵
dp2[j] = INF;↵
}↵
int fp = gcd(n - i,lst);↵
for(int j = 0; j < fp; j ++){↵
mx[j] = n + 1;↵
dp2[n + 1] = INF;↵
}↵
for(int j = 0; j < lst; j ++){↵
dp2[dp1[j] % nw] = min(dp2[dp1[j] % nw],dp1[j]);↵
if(dp2[mx[dp1[j] % fp]] > dp2[dp1[j] % nw])↵
mx[dp1[j] % fp] = dp1[j] % nw;↵
}↵
↵
for(int j = 0; j < fp; j ++){↵
long long mn = dp2[mx[j]];↵
if(mx[j] == n + 1)↵
continue;↵
for(int k = (mx[j] + lst) % nw; k != mx[j]; k = (k + lst) % nw){↵
dt[k] = -1;↵
mn = min(mn + lst,dp2[k]);↵
dp2[k] = mn;↵
}↵
}↵
// printf("%d\n",i);↵
fp = gcd(nw,diff[i]);↵
for(int j = 0; j < fp; j ++){↵
mx[j] = n + 1;↵
}↵
dp1[n + 1] = INF;↵
for(int j = 0; j < nw; j ++){↵
dt[j] = 0;↵
dp1[j] = dp2[j];↵
if(dp1[j] < dp1[mx[j % fp]])↵
mx[j % fp] = j;↵
}↵
for(int j = 0; j < fp; j ++){↵
head = 1,tail = 0;↵
for(int k = mx[j],l = 0; l < nw / fp; l ++,k = (k + diff[i]) % nw){↵
dt[k] = l;↵
while(head <= tail && l - dt[q[head]] > ls)↵
head ++;↵
if(head <= tail)↵
dp1[k] = min(dp1[k],dp1[q[head]] + 1ll * l * diff[i] - 1ll * dt[q[head]] * diff[i] + nw);↵
while(head <= tail && dp1[k] - 1ll * l * diff[i] <= dp1[q[tail]] - 1ll * dt[q[tail]] * diff[i])↵
tail --;↵
q[++tail] = k;↵
//printf("%d %lld\n",k,dp1[k]);↵
}↵
}↵
lst = nw;↵
}↵
long long ans = 0;↵
for(int i = 0; i < lst; i ++){↵
//printf("%d %lld\n",i,dp1[i]);↵
if(dp1[i] <= w)↵
ans = ans + (w - dp1[i]) / lst + 1;↵
}↵
printf("%lld\n",ans);↵
}↵
}↵
~~~~~↵
↵
</spoiler>↵
↵
# Extention to palindromes↵
↵
As we'll found that The borders of palindormes are also the palindromes suffixes of a palindrome, then we can say that for a string $S$, the lengths of palindrome suffixes form $\operatorname{O}(\log |S|)$ continuous arthmetic substrings. We can also say that even lengths of palindrome suffixes form $\operatorname{O}(\log |S|)$ continuous arthmetic substrings.↵
↵
# Shlink pointer ↵
↵
## introduction↵
↵
As $shlink_i$ means to a node $i$,which node is the start of the next arthmetic sequence,can be calculated like this:↵
↵
↵
~~~~~↵
shlink[i] = link[i];↵
diff[i] = len[i] - len[link[i]];↵
if(diff[i] == diff[link[i]])↵
shlink[i] = shlink[link[i]];↵
~~~~~↵
↵
↵
Then we'll find:↵
↵
↵
↵
Which is a good way to make use of Border Theory.↵
↵
## Tasks↵
↵
In [problem:932G] we can make a string $T$ by:↵
↵
- $t_{2k - 1} = s_{k}$↵
- $t_{2k} = s_{n - k + 1}$↵
↵
Then we found that The task turns to how to divide $T$ into several even palindroms, we tries to solve it using dynamic planning as $dp_i$ is the answer of the prefix of length $i$, then we use shlink pointers to make it $\operatorname{O}(|S| \log |S|)$.↵
↵
<spoiler summary="code">↵
↵
~~~~~↵
#include<bits/stdc++.h>↵
using namespace std;↵
int n,b[1000009],tp[1000009],to[1000009][26],len[1000009],cnt,dp[1000009],f[1000009],shlink[1000009],diff[1000009];↵
char c[1000009],s[1000009];↵
const int mod = 1000000007;↵
int main(){↵
scanf(" %s",c);↵
n = strlen(c);↵
for(int i = 0; (i + 1) << 1 <= n; i ++){↵
s[i << 1 | 1] = c[i]; ↵
}↵
for(int i = n - 1; (i + 1) << 1 > n; i --){↵
s[(n - i) << 1] = c[i];↵
}↵
//printf("%s\n",s + 1);↵
for(int i = 2; i <= n; i ++){↵
int t = tp[i - 1];↵
while(t != 0 && s[i] != s[i - 1 - len[t]])↵
t = b[t];↵
if(s[i] == s[i - 1 - len[t]] && !to[t][s[i] - 'a']){↵
cnt ++;↵
//printf("%d\n",cnt);↵
to[t][s[i] - 'a'] = cnt;↵
len[cnt] = len[t] + 2;↵
int st = b[t];↵
while(st != 0 && s[i] != s[i - 1 - len[st]])↵
st = b[st];↵
if(s[i] == s[i - 1 - len[st]] && len[cnt] != 2)↵
st = to[st][s[i] - 'a'];↵
b[cnt] = st;↵
t = cnt;↵
}↵
else if(s[i] == s[i - 1 - len[t]])↵
t = to[t][s[i] - 'a'];↵
tp[i] = t;↵
//printf("%d %d %d\n",i,len[tp[i]],len[b[tp[i]]]);↵
}↵
for(int i = 1; i <= cnt; i ++){↵
diff[i] = len[i] - len[b[i]];↵
shlink[i] = b[i];↵
if(diff[i] == diff[b[i]])↵
shlink[i] = shlink[b[i]];↵
}↵
dp[0] = 1;↵
for(int i = 2; i <= n; i += 2){↵
if(tp[i] == 0)↵
continue;↵
int p = tp[i];↵
//dp[i] = dp[i - tp[i]];↵
while(p != 0){↵
f[p] = ((shlink[p] != b[p]) * f[b[p]] + dp[i - len[shlink[p]] - diff[p]]) % mod;↵
dp[i] = (dp[i] + f[p]) % mod;↵
//printf("%d %d %d %d\n",len[p],len[shlink[p]] + diff[p],f[p],f[b[p]]);↵
p = shlink[p];↵
}↵
//printf("%d %d\n",i,dp[i]);↵
}↵
printf("%d\n",dp[n]);↵
}↵
~~~~~↵
↵
↵
</spoiler>↵
↵
Such solution is similar to [problem:906E],which turns:↵
↵
- $t_{2k - 1} = x_k$↵
- $t_{2k} = y_k$↵
↵
And use dynamic planning↵
↵
<spoiler summary="code">↵
↵
~~~~~↵
#include<bits/stdc++.h>↵
using namespace std;↵
int n,m,shlink[1000009],b[1000009],to[1000009][26],tp[1000009],len[1000009],diff[1000009],cnt;↵
char s[500009],t[500009],ct[500009];↵
int dp[1000009],las[1000009],mn[1000009],flas[1000009];↵
int main(){↵
scanf(" %s %s",s,t);↵
n = strlen(s) << 1;↵
for(int i = 0; s[i] != '\0'; i ++)↵
ct[i << 1 | 1] = s[i],ct[(i << 1) + 2] = t[i];↵
for(int i = 2; i <= n; i ++){↵
int o = tp[i - 1];↵
while(o != 0 && ct[i] != ct[i - len[o] - 1])↵
o = b[o];↵
if(ct[i] == ct[i - len[o] - 1]){↵
if(to[o][ct[i] - 'a'] == 0){↵
cnt ++;↵
to[o][ct[i] - 'a'] = cnt;↵
len[cnt] = len[o] + 2;↵
int u = b[o];↵
while(u != 0 && ct[i] != ct[i - len[u] - 1])↵
u = b[u];↵
if(to[u][ct[i] - 'a'] != cnt && ct[i] == ct[i - len[u] - 1])↵
b[cnt] = to[u][ct[i] - 'a'];↵
shlink[cnt] = b[cnt];↵
diff[cnt] = len[cnt] - len[b[cnt]];↵
if(diff[cnt] == diff[b[cnt]])↵
shlink[cnt] = shlink[b[cnt]];↵
} ↵
o = to[o][ct[i] - 'a'];↵
}↵
tp[i] = o;↵
//printf("%d %d %d\n",i,len[tp[i]],len[b[tp[i]]]);↵
}↵
mn[0] = (1 << 24);↵
for(int i = 2; i <= n; i += 2){↵
if(ct[i] == ct[i - 1]){↵
dp[i] = dp[i - 2];↵
las[i] = i - 2;↵
}↵
else{↵
dp[i] = (1 << 24);↵
}↵
int p = tp[i];↵
if(len[p] != 0){↵
do{↵
mn[p] = dp[i - len[shlink[p]] - diff[p]] + 1;↵
flas[p] = i - len[shlink[p]] - diff[p];↵
if(b[p] != shlink[p]){↵
mn[p] = min(mn[b[p]],mn[p]); ↵
if(mn[p] == mn[b[p]])↵
flas[p] = flas[b[p]];↵
}↵
dp[i] = min(dp[i],mn[p]);↵
if(dp[i] == mn[p])↵
las[i] = flas[p];↵
p = shlink[p];↵
}while(len[p] != 0);↵
}↵
//printf("%d %d %d\n",i,dp[i],las[i]);↵
}↵
if((dp[n] >> 24) & 1)↵
puts("-1");↵
else{↵
printf("%d\n",dp[n]);↵
int o = n;↵
while(o != 0){↵
if(o - las[o] != 2)↵
printf("%d %d\n",(las[o] >> 1) + 1,o >> 1);↵
o = las[o];↵
}↵
}↵
} ↵
~~~~~↵
↵
</spoiler>↵
↵
# Summary↵
↵
Sometimes we found that our solution is slow because we may have to bruteforce on borders. But when using the theory,we may found that the solution sovlvable and easy instead of others that might be complex.
↵
Border Theory,which was not well-known enough,should have been publicized together with KMP or similar algorithms.But nowadays lots of people don't know such a theory exist,making them unable to solve several tasks easily or even can't get a fast solution.So it's time to complete the blog as what I will do on this.↵
↵
# Ordinary Border Theory↵
↵
## Conclution↵
↵
For a string $S$, There are a squence of strings:↵
$$b_i(0 = b_1 < b_2 < b_3 < ... < b_n = |S|,S_{1,...,b_i} = S_{n - b_i + 1,...,n})$$↵
↵
it's garenteed that there exist a way to divide the sequence into $\operatorname{O}(\log |S|)$ continuous subsequence that each subsequence forms an arithmetic sequence.↵
↵
## Proof↵
↵
First of all,we found that if there's a string $T$ as a border of $S$ and $2|T| \ge |S|$, Then the sequence of where $T$ appears in $S$ forms an arthmetic sequence.↵
↵
<spoiler summary="proof">↵
If there's pairs of nearby places where $T$ matches $(i - p,i)$ and $(i,i + q)$ that $p \neq q$,we found that:↵
↵
- $p$ is the length of a border↵
- $q$ is the length of an another border↵
- The shorter border is a border of a larger border.↵
↵
Then we will found that if $p > q$, thenwe found pair $(i - q,i)$ exists,or $(i,i + p)$ exists while $p < q$, which are not valid when we deem pairs $(i - p,i)$,$(i,i + q)$ as nearby place that match,so $p = q$ garenteed.↵
</spoiler>↵
↵
Second,We found that all lengths of border $T$ that $2|T| \ge |S|$ will form an arthmetic sequence:↵
↵
<spoiler summary="proof">↵
if there's border $i + p,i,i - q$ as no border length $l$ that $i - q < l < i$ or $i < l < i + p$,we'll found:↵
↵
- $p,q$ are the lengths of borders.↵
- $i - p$ is the length of a border of the border with length $i$.↵
- $i$ is the length of a border of the border with length $i + q$.↵
↵
We will find $i - q < i - p < i$ or $i < i + q < i + p$, making it incorrect,so $p = q$↵
</spoiler>↵
↵
So there's a way that if $p < q,p,q \text{are the length of borders}$ not in the same continuous sequence, $2p < q$ exist,So the conclution is right.↵
↵
## Task
↵
[Link](https://www.luogu.com.cn/problem/P4156)↵
↵
Given at most $5$ cases in a test case,each cases gives a string $S(1 \le |S| \le 5 \times 10 ^ 5)$ and an integer $w(1 \le w \le 10 ^ {18})$, when we have a string $T$ we can make the new string by $|T| + |S|$ or if a suffix $A$ in $T$ is also the prefix of $S$, $T - A + S$ can be formed as there's a string of length $|T| - |A| + |S|$.Find the number of distinct length of strings in $[1,w]$ can be formed.↵
↵
<spoiler summary="solution">↵
In fact,if there's a way for an addition $x$,we can form a graph of $x$ nodes and deem other additions as edges to form a graph. We'll have to run a shortest paths algorithm to get the smallest lengths with garenteed modules.↵
↵
But the naive algorithm is too slow,as borders form $\operatorname{O}(\log |S|)$ distinct arthmetic sequences,we just start updating with an arthmetic sequence for $\operatorname{O}(\log |S|)$ rounds. Every time we start with the smallest node in each cycle and can update all the nodes in the cycle using a monotonous queue(as a sliding window problem) in $\operatorname{O}(x)$.↵
↵
At first we start with a graph with $|S|$ nodes and only node $0$ with a distance $|S|$, Then we may change $x$ to $x'$,it's also easy to start form the node with smallest distance in each cycle and update it in $\operatorname{O}(x + x')$.↵
↵
At last we get an $\operatorname{O}(|S|\log |S|)$ algorithm.↵
</spoiler>↵
↵
<spoiler summary="code">↵
↵
~~~~~↵
#include<bits/stdc++.h>↵
using namespace std;↵
int t,n,bor[500009],diff[500009],dt[500009],shlink[500009],q[500009],head,tail;↵
long long w,dp1[500009],dp2[500009];↵
int mx[500009];↵
const long long INF = 1998244353999911659ll;↵
char c[500009];↵
int gcd(int x,int y){↵
return y == 0 ? x : gcd(y,x % y);↵
}↵
int main(){↵
scanf("%d",&t);↵
while(t--){↵
scanf("%d %lld",&n,&w);↵
scanf(" %s",c + 1);↵
bor[1] = 0;↵
for(int i = 2; i <= n; i ++){↵
int t = bor[i - 1];↵
while(t != 0 && c[i] != c[t + 1])↵
t = bor[t];↵
if(c[i] == c[t + 1])↵
t ++;↵
bor[i] = t;↵
}↵
for(int i = 1; i <= n; i ++){↵
diff[i] = i - bor[i];↵
shlink[i] = bor[i];↵
if(diff[i] == diff[bor[i]])↵
shlink[i] = shlink[bor[i]];↵
}↵
//for(int i = n; i > 0; i = bor[i])↵
// printf("%d\n",bor[i]);↵
//puts("OK");↵
dp1[0] = n;↵
for(int i = 1; i < n; i ++){↵
dp1[i] = INF;↵
}↵
int lst = n;↵
for(int i = bor[n]; i != 0; i = shlink[i]){↵
int nw = n - i;↵
int ls = (i - shlink[i]) / diff[i] - 1;↵
for(int j = 0; j < nw; j ++){↵
dp2[j] = INF;↵
}↵
int fp = gcd(n - i,lst);↵
for(int j = 0; j < fp; j ++){↵
mx[j] = n + 1;↵
dp2[n + 1] = INF;↵
}↵
for(int j = 0; j < lst; j ++){↵
dp2[dp1[j] % nw] = min(dp2[dp1[j] % nw],dp1[j]);↵
if(dp2[mx[dp1[j] % fp]] > dp2[dp1[j] % nw])↵
mx[dp1[j] % fp] = dp1[j] % nw;↵
}↵
↵
for(int j = 0; j < fp; j ++){↵
long long mn = dp2[mx[j]];↵
if(mx[j] == n + 1)↵
continue;↵
for(int k = (mx[j] + lst) % nw; k != mx[j]; k = (k + lst) % nw){↵
dt[k] = -1;↵
mn = min(mn + lst,dp2[k]);↵
dp2[k] = mn;↵
}↵
}↵
// printf("%d\n",i);↵
fp = gcd(nw,diff[i]);↵
for(int j = 0; j < fp; j ++){↵
mx[j] = n + 1;↵
}↵
dp1[n + 1] = INF;↵
for(int j = 0; j < nw; j ++){↵
dt[j] = 0;↵
dp1[j] = dp2[j];↵
if(dp1[j] < dp1[mx[j % fp]])↵
mx[j % fp] = j;↵
}↵
for(int j = 0; j < fp; j ++){↵
head = 1,tail = 0;↵
for(int k = mx[j],l = 0; l < nw / fp; l ++,k = (k + diff[i]) % nw){↵
dt[k] = l;↵
while(head <= tail && l - dt[q[head]] > ls)↵
head ++;↵
if(head <= tail)↵
dp1[k] = min(dp1[k],dp1[q[head]] + 1ll * l * diff[i] - 1ll * dt[q[head]] * diff[i] + nw);↵
while(head <= tail && dp1[k] - 1ll * l * diff[i] <= dp1[q[tail]] - 1ll * dt[q[tail]] * diff[i])↵
tail --;↵
q[++tail] = k;↵
//printf("%d %lld\n",k,dp1[k]);↵
}↵
}↵
lst = nw;↵
}↵
long long ans = 0;↵
for(int i = 0; i < lst; i ++){↵
//printf("%d %lld\n",i,dp1[i]);↵
if(dp1[i] <= w)↵
ans = ans + (w - dp1[i]) / lst + 1;↵
}↵
printf("%lld\n",ans);↵
}↵
}↵
~~~~~↵
↵
</spoiler>↵
↵
# Extention to palindromes↵
↵
As we'll found that The borders of palindormes are also the palindromes suffixes of a palindrome, then we can say that for a string $S$, the lengths of palindrome suffixes form $\operatorname{O}(\log |S|)$ continuous arthmetic substrings. We can also say that even lengths of palindrome suffixes form $\operatorname{O}(\log |S|)$ continuous arthmetic substrings.↵
↵
# Shlink pointer ↵
↵
## introduction↵
↵
As $shlink_i$ means to a node $i$,which node is the start of the next arthmetic sequence,can be calculated like this:↵
↵
↵
~~~~~↵
shlink[i] = link[i];↵
diff[i] = len[i] - len[link[i]];↵
if(diff[i] == diff[link[i]])↵
shlink[i] = shlink[link[i]];↵
~~~~~↵
↵
↵
Then we'll find:↵
↵
↵
↵
Which is a good way to make use of Border Theory.↵
↵
## Tasks↵
↵
In [problem:932G] we can make a string $T$ by:↵
↵
- $t_{2k - 1} = s_{k}$↵
- $t_{2k} = s_{n - k + 1}$↵
↵
Then we found that The task turns to how to divide $T$ into several even palindroms, we tries to solve it using dynamic planning as $dp_i$ is the answer of the prefix of length $i$, then we use shlink pointers to make it $\operatorname{O}(|S| \log |S|)$.↵
↵
<spoiler summary="code">↵
↵
~~~~~↵
#include<bits/stdc++.h>↵
using namespace std;↵
int n,b[1000009],tp[1000009],to[1000009][26],len[1000009],cnt,dp[1000009],f[1000009],shlink[1000009],diff[1000009];↵
char c[1000009],s[1000009];↵
const int mod = 1000000007;↵
int main(){↵
scanf(" %s",c);↵
n = strlen(c);↵
for(int i = 0; (i + 1) << 1 <= n; i ++){↵
s[i << 1 | 1] = c[i]; ↵
}↵
for(int i = n - 1; (i + 1) << 1 > n; i --){↵
s[(n - i) << 1] = c[i];↵
}↵
//printf("%s\n",s + 1);↵
for(int i = 2; i <= n; i ++){↵
int t = tp[i - 1];↵
while(t != 0 && s[i] != s[i - 1 - len[t]])↵
t = b[t];↵
if(s[i] == s[i - 1 - len[t]] && !to[t][s[i] - 'a']){↵
cnt ++;↵
//printf("%d\n",cnt);↵
to[t][s[i] - 'a'] = cnt;↵
len[cnt] = len[t] + 2;↵
int st = b[t];↵
while(st != 0 && s[i] != s[i - 1 - len[st]])↵
st = b[st];↵
if(s[i] == s[i - 1 - len[st]] && len[cnt] != 2)↵
st = to[st][s[i] - 'a'];↵
b[cnt] = st;↵
t = cnt;↵
}↵
else if(s[i] == s[i - 1 - len[t]])↵
t = to[t][s[i] - 'a'];↵
tp[i] = t;↵
//printf("%d %d %d\n",i,len[tp[i]],len[b[tp[i]]]);↵
}↵
for(int i = 1; i <= cnt; i ++){↵
diff[i] = len[i] - len[b[i]];↵
shlink[i] = b[i];↵
if(diff[i] == diff[b[i]])↵
shlink[i] = shlink[b[i]];↵
}↵
dp[0] = 1;↵
for(int i = 2; i <= n; i += 2){↵
if(tp[i] == 0)↵
continue;↵
int p = tp[i];↵
//dp[i] = dp[i - tp[i]];↵
while(p != 0){↵
f[p] = ((shlink[p] != b[p]) * f[b[p]] + dp[i - len[shlink[p]] - diff[p]]) % mod;↵
dp[i] = (dp[i] + f[p]) % mod;↵
//printf("%d %d %d %d\n",len[p],len[shlink[p]] + diff[p],f[p],f[b[p]]);↵
p = shlink[p];↵
}↵
//printf("%d %d\n",i,dp[i]);↵
}↵
printf("%d\n",dp[n]);↵
}↵
~~~~~↵
↵
↵
</spoiler>↵
↵
Such solution is similar to [problem:906E],which turns:↵
↵
- $t_{2k - 1} = x_k$↵
- $t_{2k} = y_k$↵
↵
And use dynamic planning↵
↵
<spoiler summary="code">↵
↵
~~~~~↵
#include<bits/stdc++.h>↵
using namespace std;↵
int n,m,shlink[1000009],b[1000009],to[1000009][26],tp[1000009],len[1000009],diff[1000009],cnt;↵
char s[500009],t[500009],ct[500009];↵
int dp[1000009],las[1000009],mn[1000009],flas[1000009];↵
int main(){↵
scanf(" %s %s",s,t);↵
n = strlen(s) << 1;↵
for(int i = 0; s[i] != '\0'; i ++)↵
ct[i << 1 | 1] = s[i],ct[(i << 1) + 2] = t[i];↵
for(int i = 2; i <= n; i ++){↵
int o = tp[i - 1];↵
while(o != 0 && ct[i] != ct[i - len[o] - 1])↵
o = b[o];↵
if(ct[i] == ct[i - len[o] - 1]){↵
if(to[o][ct[i] - 'a'] == 0){↵
cnt ++;↵
to[o][ct[i] - 'a'] = cnt;↵
len[cnt] = len[o] + 2;↵
int u = b[o];↵
while(u != 0 && ct[i] != ct[i - len[u] - 1])↵
u = b[u];↵
if(to[u][ct[i] - 'a'] != cnt && ct[i] == ct[i - len[u] - 1])↵
b[cnt] = to[u][ct[i] - 'a'];↵
shlink[cnt] = b[cnt];↵
diff[cnt] = len[cnt] - len[b[cnt]];↵
if(diff[cnt] == diff[b[cnt]])↵
shlink[cnt] = shlink[b[cnt]];↵
} ↵
o = to[o][ct[i] - 'a'];↵
}↵
tp[i] = o;↵
//printf("%d %d %d\n",i,len[tp[i]],len[b[tp[i]]]);↵
}↵
mn[0] = (1 << 24);↵
for(int i = 2; i <= n; i += 2){↵
if(ct[i] == ct[i - 1]){↵
dp[i] = dp[i - 2];↵
las[i] = i - 2;↵
}↵
else{↵
dp[i] = (1 << 24);↵
}↵
int p = tp[i];↵
if(len[p] != 0){↵
do{↵
mn[p] = dp[i - len[shlink[p]] - diff[p]] + 1;↵
flas[p] = i - len[shlink[p]] - diff[p];↵
if(b[p] != shlink[p]){↵
mn[p] = min(mn[b[p]],mn[p]); ↵
if(mn[p] == mn[b[p]])↵
flas[p] = flas[b[p]];↵
}↵
dp[i] = min(dp[i],mn[p]);↵
if(dp[i] == mn[p])↵
las[i] = flas[p];↵
p = shlink[p];↵
}while(len[p] != 0);↵
}↵
//printf("%d %d %d\n",i,dp[i],las[i]);↵
}↵
if((dp[n] >> 24) & 1)↵
puts("-1");↵
else{↵
printf("%d\n",dp[n]);↵
int o = n;↵
while(o != 0){↵
if(o - las[o] != 2)↵
printf("%d %d\n",(las[o] >> 1) + 1,o >> 1);↵
o = las[o];↵
}↵
}↵
} ↵
~~~~~↵
↵
</spoiler>↵
↵
# Summary↵
↵
Sometimes we found that our solution is slow because we may have to bruteforce on borders. But when using the theory,we may found that the solution sovlvable and easy instead of others that might be complex.
