HDU 5558 Alice's Classified Message(后缀数组+二分+rmq(+线段树?))
题意
大概就是给你一个串，对于每个\(i\)，在\([1,i-1]\)中找到一个\(j\)，使得\(lcp(i,j)\)最长，若有多个最大\(j\)选最小，求\(j\)和这个\(lcp\)长度
思路
首先我们需要知道对于每个\(i\)，能与下标小于\(i\)开头的前缀构成的最大\(lcp\)是多少
这个可以在最外层枚举\(i\)的过程中维护一个\(set\)，这样在插入当前的\(rk[i]\)的时候能\(O(logn)\)得到这个最长的\(lcp\)
然后根据这个值二分出\(rk[i]\)向左右能扩展的最远的地方\([L,R]\)，使得\(lcp\)依然为这个值
\([L,R]\)内的最小的\(sa\)就是答案
脑子抽了用了个线段树维护已经存在过的下标，没出现的用\(inf\)表示，但实际上无论是inf还是本身它都不可能是真正的答案，所以这边还是rmq
不过复杂度不影响，少了点常数..
代码
下面注释提供了一些自己拍出来的数据
谁让我不会自动机呢
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>
#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
using namespace std;
typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;
typedef pair<ll,int> PIL;
const db eps = 1e-6;
const int mod = 1e9+7;
const int maxn = 3e5+100;
const int maxm = 2e6+100;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);
int t;
int n,m;
char s[maxn];
int sa[maxn],rk[maxn],height[maxn];
int y[maxn],x[maxn],c[maxn];
void getSa(){
for(int i=1;i<=n;i++)++c[x[i]=s[i]];
for(int i=2;i<=m;i++)c[i]+=c[i-1];
for(int i=n;i>=1;i--){
//printf("sa[%d]=%d\n",c[x[i]],i);
sa[c[x[i]]--]=i;
}
for(int k=1;k<=n;k<<=1){
//printf("^^^^%d\n",k);
int num = 0;
for(int i=n-k+1;i<=n;i++)y[++num]=i;
for(int i=1;i<=n;i++)if(sa[i]>k)y[++num]=sa[i]-k;
for(int i=1;i<=m;i++)c[i]=0;
for(int i=1;i<=n;i++)++c[x[i]];
for(int i=2;i<=m;i++)c[i]+=c[i-1];
for(int i=n;i>=1;i--)sa[c[x[y[i]]]--]=y[i],y[i]=0;
swap(x,y);
x[sa[1]]=1;
num=1;
for(int i = 1; i <= n; i++){
//printf("y[%d]=%d\n",i,y[i]);
}
for(int i=2;i<=n;i++){
x[sa[i]]=(y[sa[i]]==y[sa[i-1]]&&y[sa[i]+k]==y[sa[i-1]+k])?num:++num;
/*printf("num==%d\n",num);
printf("x[%d]=(y[%d]==y[%d]&&y[%d]==y[%d])?num:++num\n",sa[i],sa[i],sa[i-1],sa[i]+k,sa[i-1]+k);*/
}
if(num==n)break;
m=num;
}
}
void getHeight(){
int k=0;
for(int i=1; i<=n; ++i)rk[sa[i]]=i;
for(int i=1; i<=n; ++i){
if(rk[i]==1) continue;
if(k)--k;
int j=sa[rk[i]-1];
while(j+k<=n&&i+k<=n&&s[i+k]==s[j+k])++k;
height[rk[i]]=k;
}
}
set<int>S;
int d[maxn][23];
void init(){
for(int i = 1; i <= n; i++)d[i][0]=height[i];
for(int j = 1; (1<<j)<=n; j++){
for(int i = 1; i+(1<<j)-1<=n; i++){
d[i][j]=min(d[i][j-1],d[i+(1<<(j-1))][j-1]);
}
}
}
int rmq(int l, int r){
if (l > r) swap (l, r);
int k = 0;
while((1<<(k+1))<=r-l+1)k++;
return min(d[l][k],d[r-(1<<k)+1][k]);
}
int mi[maxn<<2];
void build(int l, int r, int root){
int mid = l+r>>1;
if(l==r){mi[root]=inf;return;}
build(lson);build(rson);
mi[root]=min(mi[lc],mi[rc]);
return;
}
void update(int x, int y, int l, int r, int root){
int mid = l+r>>1;
if(l==r){mi[root]=y;return;}
if(x<=mid)update(x,y,lson);
else update(x,y,rson);
mi[root]=min(mi[lc],mi[rc]);
return;
}
int ask(int x, int y, int l, int r, int root){
int mid=l+r>>1;
if(x<=l&&r<=y)return mi[root];
int ans = inf;
if(x<=mid)ans=min(ans,ask(x,y,lson));
if(y>mid)ans=min(ans,ask(x,y,rson));
return ans;
}
int ntot;
PI Ans[maxn];
int main(){
scanf("%d", &t);
int ncase = 0;
while(t--){
scanf("%s",s+1);
n = strlen(s+1);
m=122;
for(int i = 0; i <= m; i++)c[i]=0;
//for(int i = 0; i <= 2*n+2; i++)x[i]=y[i]=0;
getSa();
/*
for(int i = 1; i <= n; i++){
printf(" %d %d %d\n",i,x[i],y[i]);
}*/
getHeight();
printf("Case #%d:\n",++ncase);
S.clear();
init();
build(1,n,1);
for(int i = 1; i <= n; ){
int mxLen = 0;
set<int>::iterator it = S.lower_bound(rk[i]);
if(it!=S.end()){
mxLen=max(mxLen,rmq(rk[i]+1,(*it)));
}
if(it!=S.begin()){
it--;
mxLen=max(mxLen,rmq((*it)+1,rk[i]));
}
S.insert(rk[i]);
if(!mxLen){
printf("-1 %d\n",s[i]);
update(rk[i],i,1,n,1);
i++;
continue;
}
int L = -1,R=-1;
int l=1,r=rk[i]-1;
while(l<=r){
int mid=l+r>>1;
if(rmq(mid+1,rk[i])>=mxLen){
L=mid;r=mid-1;
}
else l=mid+1;
}
l=rk[i]+1,r=n;
while(l<=r){
int mid=l+r>>1;
int tmp = rmq(rk[i]+1,mid);
if(rmq(rk[i]+1,mid)>=mxLen){
R=mid;l=mid+1;
}
else r=mid-1;
}
int to = i+mxLen;
if(L==-1)L=inf;
else L=ask(L,rk[i],1,n,1);
if(R==-1)R=inf;
else R=ask(rk[i],R,1,n,1);
int ans=min(L,R);
printf("%d %d\n",mxLen,ans-1);
while(i<to){
S.insert(rk[i]);
update(rk[i],i,1,n,1);
i++;
if(i>n)break;
}
}
}
return 0;
}
/*
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*/
