问题：
The little cat is majoring in physics in the capital of Byterland. A piece of sad news comes to him these days: his mother is getting ill. Being worried about spending so much on railway tickets (Byterland is such a big country, and he has to spend 16 shours on train to his hometown), he decided only to send SMS with his mother.
The little cat lives in an unrich family, so he frequently comes to the mobile service center, to check how much money he has spent on SMS. Yesterday, the computer of service center was broken, and printed two very long messages. The brilliant little cat soon found out:
1. All characters in messages are lowercase Latin letters, without punctuations and spaces.
2. All SMS has been appended to each other – (i+1)-th SMS comes directly after the i-th one – that is why those two messages are quite long.
3. His own SMS has been appended together, but possibly a great many redundancy characters appear leftwards and rightwards due to the broken computer.
E.g: if his SMS is “motheriloveyou”, either long message printed by that machine, would possibly be one of “hahamotheriloveyou”, “motheriloveyoureally”, “motheriloveyouornot”, “bbbmotheriloveyouaaa”, etc.
4. For these broken issues, the little cat has printed his original text twice (so there appears two very long messages). Even though the original text remains the same in two printed messages, the redundancy characters on both sides would be possibly different.
You are given those two very long messages, and you have to output the length of the longest possible original text written by the little cat.
Background:
The SMS in Byterland mobile service are charging in dollars-per-byte. That is why the little cat is worrying about how long could the longest original text be.
Why ask you to write a program? There are four resions:
1. The little cat is so busy these days with physics lessons;
2. The little cat wants to keep what he said to his mother seceret;
3. POJ is such a great Online Judge;
4. The little cat wants to earn some money from POJ, and try to persuade his mother to see the doctor :(
Input
Two strings with lowercase letters on two of the input lines individually. Number of characters in each one will never exceed 100000.
Output
A single line with a single integer number – what is the maximum length of the original text written by the little cat.
Sample Input
yeshowmuchiloveyoumydearmotherreallyicannotbelieveit yeaphowmuchiloveyoumydearmother
Sample Output
27
题意：
求两个字符串的最长的公共字串。
思路：
后缀自动机：可以直接匹配，然后又默写了一遍后缀自动机。
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=3000000;
char chr[maxn],str[maxn];
struct SAM
{
int ch[maxn][26],fa[maxn],maxlen[maxn],Last,sz;
void init()
{
sz=Last=1; fa[1]=maxlen[1]=0;
memset(ch[1],0,sizeof(ch[1]));
}
void add(int x)
{
int np=++sz,p=Last;Last=np;
memset(ch[np],0,sizeof(ch[np]));
maxlen[np]=maxlen[p]+1;
while(p&&!ch[p][x]) ch[p][x]=np,p=fa[p];
if(!p) fa[np]=1;
else {
int q=ch[p][x];
if(maxlen[p]+1==maxlen[q]) fa[np]=q;
else {
int nq=++sz;
memcpy(ch[nq],ch[q],sizeof(ch[q]));
maxlen[nq]=maxlen[p]+1;
fa[nq]=fa[q];
fa[q]=fa[np]=nq;
while(p&&ch[p][x]==q) ch[p][x]=nq,p=fa[p];
}
}
}
void solve()
{
scanf("%s",chr);
int L=strlen(chr),x,tmp=0,ans=0;Last=1;
for(int i=0;i<L;i++){
x=chr[i]-‘a‘;
if(ch[Last][x]) tmp++,Last=ch[Last][x];
else {
while(Last&&!ch[Last][x]) Last=fa[Last];
if(!Last) tmp=0,Last=1;
else tmp=maxlen[Last]+1,Last=ch[Last][x];
}
ans=max(ans,tmp);
}
printf("%d\n",ans);
}
};
SAM Sam;
int main()
{
int T,i,L;
while(~scanf("%s",chr)){
Sam.init();
L=strlen(chr);
for(i=0;i<L;i++) Sam.add(chr[i]-‘a‘);
Sam.solve();
}
return 0;
}
后缀数组：需要把两个串连接起来，之间加一个特殊符号，用来保证得到的结果符合两个串来自不同的母串。
效率对比：
后缀自动机94ms，后缀数组782ms。这种基本题型我还是愿意写后缀自动机，不过练一练总是有好处的。
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=4100000;
char str1[maxn],str2[maxn];
int L,ch[maxn];
struct SA
{
int cntA[maxn],cntB[maxn],A[maxn],B[maxn];
int rank[maxn],sa[maxn],tsa[maxn],ht[maxn];
void sort()
{
for (int i = 0; i < 28; i ++) cntA[i] = 0;
for (int i = 1; i <= L; i ++) cntA[ch[i]] ++;
for (int i = 1; i < 28; i ++) cntA[i] += cntA[i - 1];
for (int i = L; i; i --) sa[cntA[ch[i]] --] = i;
rank[sa[1]] = 1;
for (int i = 2; i <= L; i ++){
rank[sa[i]] = rank[sa[i - 1]];
if (ch[sa[i]] != ch[sa[i - 1]]) rank[sa[i]] ++;
}
for (int l = 1; rank[sa[L]] < L; l <<= 1){
for (int i = 0; i <= L; i ++) cntA[i] = 0;
for (int i = 0; i <= L; i ++) cntB[i] = 0;
for ( int i = 1; i <= L; i ++){
cntA[A[i] = rank[i]] ++;
cntB[B[i] = (i + l <= L) ? rank[i + l] : 0] ++;
}
for (int i = 1; i <= L; i ++) cntB[i] += cntB[i - 1];
for (int i = L; i; i --) tsa[cntB[B[i]] --] = i;
for (int i = 1; i <= L; i ++) cntA[i] += cntA[i - 1];
for (int i = L; i; i --) sa[cntA[A[tsa[i]]] --] = tsa[i];
rank[sa[1]] = 1;
for (int i = 2; i <= L; i ++){
rank[sa[i]] = rank[sa[i - 1]];
if (A[sa[i]] != A[sa[i - 1]] || B[sa[i]] != B[sa[i - 1]]) rank[sa[i]] ++;
}
}
}
void getht()
{
for (int i = 1, j = 0; i <= L; i ++){
if (j) j --;
while (ch[i + j] == ch[sa[rank[i] - 1] + j]) j ++;
ht[rank[i]] = j;
}
}
};
SA Sa;
int main()
{
scanf("%s",str1+1);
scanf("%s",str2+1);
int L1=strlen(str1+1);
int L2=strlen(str2+1);
for(int i=1;i<=L1;i++) ch[i]=str1[i]-‘a‘+1;
ch[L1+1]=27;
for(int i=1;i<=L2;i++) ch[i+L1+1]=str2[i]-‘a‘+1;
L=L1+L2+1;
Sa.sort();
Sa.getht();
int ans=0;
for(int i = 1; i <= L; i++)
{
if((Sa.sa[i]<=L1)!=(Sa.sa[i-1]<=L1))
ans = max(ans, Sa.ht[i]);
}
printf("%d\n",ans);
return 0;
}