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
",c[x[i]],i);
        sa[c[x[i]]--]=i;
    }
    for(int k=1;k<=n;k<<=1){
        //printf("^^^^%d
",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
",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
",num);
            printf("x[%d]=(y[%d]==y[%d]&&y[%d]==y[%d])?num:++num
",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
",i,x[i],y[i]);
        }*/
        getHeight();
        printf("Case #%d:
",++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
",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
",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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