题意:求两个串的最长公共子串
两个串连接起来然后求高度数组
注意两个sa值必须分别在不同一侧
本题是用来测试模板的,回想起青岛那次翻车感觉很糟糕
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define rrep(i,j,k) for(register int i=j;i>=k;i--)
#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])
#define iin(a) scanf("%d",&a)
#define lin(a) scanf("%lld",&a)
#define din(a) scanf("%lf",&a)
#define s0(a) scanf("%s",a)
#define s1(a) scanf("%s",a+1)
#define print(a) printf("%lld",(ll)a)
#define enter putchar(‘\n‘)
#define blank putchar(‘ ‘)
#define println(a) printf("%lld\n",(ll)a)
#define IOS ios::sync_with_stdio(0)
using namespace std;
const int maxn = 1e6+11;
const int oo = 0x3f3f3f3f;
const double eps = 1e-7;
typedef long long ll;
ll read(){
ll x=0,f=1;register char ch=getchar();
while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
return x*f;
}
char str[maxn];
struct SA{
int Rank[maxn],sa[maxn],tsa[maxn],A[maxn],B[maxn];
int cntA[maxn],cntB[maxn];
int height[maxn],best[maxn][30],n;//height[i]:第sa[i]与sa[i-1]的cp
void get(){
n=strlen(str+1);
rep(i,0,300) cntA[i]=0;
rep(i,1,n) cntA[str[i]]++;
rep(i,1,300) cntA[i]+=cntA[i-1];
rrep(i,n,1) sa[cntA[str[i]]--]=i;
Rank[sa[1]]=1;
rep(i,2,n){
if(str[sa[i]]==str[sa[i-1]]){
Rank[sa[i]]=Rank[sa[i-1]];
}else{
Rank[sa[i]]=1+Rank[sa[i-1]];
}
}
for(int l=1;Rank[sa[n]]<n;l<<=1){
rep(i,1,n) cntA[i]=cntB[i]=0;
rep(i,1,n) cntA[A[i]=Rank[i]]++;
rep(i,1,n) cntB[B[i]=(i+l<=n?Rank[i+l]:0)]++;
rep(i,1,n) cntA[i]+=cntA[i-1],cntB[i]+=cntB[i-1];
rrep(i,n,1) tsa[cntB[B[i]]--]=i;
rrep(i,n,1) sa[cntA[A[tsa[i]]]--]=tsa[i];
Rank[sa[1]]=1;
rep(i,2,n){
bool flag=A[sa[i]]==A[sa[i-1]]&&B[sa[i]]==B[sa[i-1]];
flag=!flag;
Rank[sa[i]]=Rank[sa[i-1]]+flag;
}
}
}
void ht(){
int j=0;
rep(i,1,n){
if(j) j--;
while(str[i+j]==str[sa[Rank[i]-1]+j]) j++;
height[Rank[i]]=j;
}
}
void rmq(){
rep(i,1,n) best[i][0]=height[i];
for(int i=1;(1<<i)<=n;i++){
for(int j=1;j+(1<<i)-1<=n;j++){
best[j][i]=min(best[j][i],best[j+(1<<(i-1))][i-1]);
}
}
}
int query(int l,int r){
if(l==r)return -oo;
if(l>r)swap(l,r);
l++;
int k=log2(r-l+1);
return min(best[l][k],best[r-(1<<k)+1][k]);
}
}sa;
char str1[maxn];
int main(){
while(~s1(str)){
int len=strlen(str+1);
s1(str1);
strcat(str+1,str1+1);
sa.get();
sa.ht();
int n=strlen(str+1);
int ans=0;
for(int i = 1; i <= n; i++){
if(ans<sa.height[i]){
if((sa.sa[i]<=len&&sa.sa[i-1]>len)||(sa.sa[i]>len&&sa.sa[i-1]<=len)){
ans=sa.height[i];
}
}
}
println(ans);
}
return 0;
}