将字符串s的所有后缀排序
#include<algorithm>
#include<iostream>
#include<cstring>
#include<iomanip>
#include<sstream>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<cmath>
#include<stack>
#include<set>
#include<map>
#define rep(i,x,n) for(int i=x;i<=n;++i)
#define per(i,n,x) for(int i=n;i>=x;--i)
#define sz(a) int(a.size())
#define rson mid+1,r,p<<1|1
#define pii pair<int,int>
#define lson l,mid,p<<1
#define ll long long
#define pb push_back
#define mp make_pair
#define se second
#define fi first
using namespace std;
const double eps=1e-8;
const int mod=1e9+7;
const int N=1e6+10;
const int inf=1e9;
char s[N];
int n;
int sa[N],rk[N],wb[N],c[N];
void bd(int m){
int *x=rk,*y=wb;
rep(i,1,m) c[i]=0;
rep(i,1,n) ++c[ x[i]=s[i] ];
rep(i,2,m) c[i]+=c[i-1];
per(i,n,1) sa[ c[x[i]]-- ]=i;
for(int k=1;k<=n;k<<=1){
int p=0;
rep(i,n-k+1,n) y[++p]=i;
rep(i,1,n) if(sa[i]>k) y[++p]=sa[i]-k;
rep(i,1,m) c[i]=0;
rep(i,1,n) ++c[ x[y[i]] ];
rep(i,2,m) c[i]+=c[i-1];
per(i,n,1) sa[ c[x[y[i]]]-- ]=y[i],y[i]=0;
swap(x,y);
x[sa[1]]=1;
p=1;
rep(i,2,n)
x[sa[i]] = (y[sa[i]]==y[sa[i-1]]&&y[sa[i]+k]==y[sa[i-1]+k]) ? p : ++p;
if(p>=n) break;
m=p;
}
}
int main(){
//ios::sync_with_stdio(false);
//freopen("in","r",stdin);
scanf("%s",s+1);
n=strlen(s+1);
bd(256);
rep(i,1,n){
printf("%d%c",sa[i],i==n?'
':' ');
}
return 0;
}
查询字符串t在字符串s中出现的所有位置
#include<algorithm>
#include<iostream>
#include<cstring>
#include<iomanip>
#include<sstream>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<cmath>
#include<stack>
#include<set>
#include<map>
#define rep(i,x,n) for(int i=x;i<=n;++i)
#define per(i,n,x) for(int i=n;i>=x;--i)
#define sz(a) int(a.size())
#define rson mid+1,r,p<<1|1
#define pii pair<int,int>
#define lson l,mid,p<<1
#define ll long long
#define pb push_back
#define mp make_pair
#define se second
#define fi first
using namespace std;
const double eps=1e-8;
const int mod=1e9+7;
const int N=1e6+10;
const int inf=1e9;
char s[N],t[N];
int n,m;
int sa[N],rk[N],wb[N],c[N],nex[N];
void bd(int m){
int *x=rk,*y=wb;
rep(i,1,m) c[i]=0;
rep(i,1,n) ++c[ x[i]=s[i] ];
rep(i,2,m) c[i]+=c[i-1];
per(i,n,1) sa[ c[x[i]]-- ]=i;
for(int k=1;k<=n;k<<=1){
int p=0;
rep(i,n-k+1,n) y[++p]=i;
rep(i,1,n) if(sa[i]>k) y[++p]=sa[i]-k;
rep(i,1,m) c[i]=0;
rep(i,1,n) ++c[ x[y[i]] ];
rep(i,2,m) c[i]+=c[i-1];
per(i,n,1) sa[ c[x[y[i]]]-- ]=y[i],y[i]=0;
swap(x,y);
x[sa[1]]=1;
p=1;
rep(i,2,n)
x[sa[i]] = (y[sa[i]]==y[sa[i-1]]&&y[sa[i]+k]==y[sa[i-1]+k]) ? p : ++p;
if(p>=n) break;
m=p;
}
}
void get(char *s){
int n=strlen(s+1);
for(int i=2,j=0;i<=n;i++){
while(j&&s[i]!=s[j+1]) j=nex[j];
if(s[i]==s[j+1]) ++j;
nex[i]=j;
}
}
int main(){
//ios::sync_with_stdio(false);
//freopen("in","r",stdin);
scanf("%s",s+1);
n=strlen(s+1);
bd(256);
scanf("%s",t+1);
m=strlen(t+1);
int l=1,r=n,L=n+1,R=0;
while(l<=r){
int mid=l+r>>1;
int res=strncmp(t+1,s+sa[mid],m);
if(res<=0) r=mid-1;
else l=mid+1;
}
L=l;
l=1,r=n;
while(l<=r){
int mid=l+r>>1;
int res=strncmp(t+1,s+sa[mid],m);
if(res>=0) l=mid+1;
else r=mid-1;
}
R=r;
rep(i,1,n) c[i]=0;
rep(i,L,R) ++c[sa[i]];
rep(i,1,n) if(c[i]) printf("%d
",i);
get(t);
rep(i,1,m) printf("%d%c",nex[i],i==n?'
':' ');
return 0;
}
字符串s的任意两个后缀的最长公共前缀LCP
利用引理(height[rk[i]]ge height[rk[i-1]]-1)来(O(n))求出(height)数组。
#include<algorithm>
#include<iostream>
#include<cstring>
#include<iomanip>
#include<sstream>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<cmath>
#include<stack>
#include<set>
#include<map>
#define rep(i,x,n) for(int i=x;i<=n;++i)
#define per(i,n,x) for(int i=n;i>=x;--i)
#define sz(a) int(a.size())
#define rson mid+1,r,p<<1|1
#define pii pair<int,int>
#define lson l,mid,p<<1
#define ll long long
#define pb push_back
#define mp make_pair
#define se second
#define fi first
using namespace std;
const double eps=1e-8;
const int mod=1e9+7;
const int N=1e6+10;
const int inf=1e9;
char s[N],t[N];
int n,m;
int sa[N],rk[N],wb[N],c[N],ht[N],st[N][20];
void bd(int m){
int *x=rk,*y=wb;
rep(i,1,m) c[i]=0;
rep(i,1,n) ++c[ x[i]=s[i] ];
rep(i,2,m) c[i]+=c[i-1];
per(i,n,1) sa[ c[x[i]]-- ]=i;
for(int k=1;k<=n;k<<=1){
int p=0;
rep(i,n-k+1,n) y[++p]=i;
rep(i,1,n) if(sa[i]>k) y[++p]=sa[i]-k;
rep(i,1,m) c[i]=0;
rep(i,1,n) ++c[ x[y[i]] ];
rep(i,2,m) c[i]+=c[i-1];
per(i,n,1) sa[ c[x[y[i]]]-- ]=y[i],y[i]=0;
swap(x,y);
x[sa[1]]=1;
p=1;
rep(i,2,n)
x[sa[i]] = (y[sa[i]]==y[sa[i-1]]&&y[sa[i]+k]==y[sa[i-1]+k]) ? p : ++p;
if(p>=n) break;
m=p;
}
}
void get(){
for(int i=1,k=0;i<=n;i++){
if(k) --k;
while(s[i+k]==s[ sa[rk[i]-1] ]+k) ++k;
ht[rk[i]]=k;
}
rep(i,1,n) st[i][0]=ht[i];
for(int j=1;(1<<j)<=n;j++){
for(int i=1;i+(1<<j)-1<=n;i++){
st[i][j]=min(st[i][j-1],st[i+1<<(j-1)][j-1]);
}
}
}
int lcp(int l,int r){
int k=(int)log2(r-l+1);
return min(st[l][k],st[r-(1<<k)+1][k]);
}
int main(){
//ios::sync_with_stdio(false);
//freopen("in","r",stdin);
scanf("%s",s+1);
n=strlen(s+1);
bd(256);
get();
return 0;
}
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