BZOJ 1692: [Usaco2007 Dec]队列变换 (后缀数组/二分+Hash)

跟BZOJ 4278: [ONTAK2015]Tasowanie一模一样

SA的做法就是把原串倒过来接在原串后面,$O(nlogn)$做后缀数组,就能$O(1)$够比较每个前缀和后缀谁的字典序小了.

Hash二分也可以.

CODE(SA)

$O(nlogn)$

#include<bits/stdc++.h>
using namespace std;
char cb[1<<15],*cs=cb,*ct=cb;
#define getc() (cs==ct&&(ct=(cs=cb)+fread(cb,1,1<<15,stdin),cs==ct)?0:*cs++)
template<class T>inline void read(T &res) {
	char ch; int flg = 1; while(!isdigit(ch=getchar()))if(ch=='-')flg=-flg;
	for(res=ch-'0';isdigit(ch=getchar());res=res*10+ch-'0'); res*=flg;
}
const int MAXN = 60005;
int s[MAXN];
int x[MAXN], y[MAXN], c[MAXN], rk[MAXN], sa[MAXN];
inline void Get_Sa(int n, int m) {
    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) sa[c[x[i]]--] = i;
    for(int k = 1; k <= n; k<<=1) {
        int p = 0;
        for(int i = n-k+1; i <= n; ++i) y[++p] = i;
        for(int i = 1; i <= n; ++i) if(sa[i]>k) y[++p] = 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];
        swap(x, y);
        x[sa[1]] = p = 1;
        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]) ? p : ++p;
        if((m=p) == n) break;
    }
    for(int i = 1; i <= n; ++i) rk[sa[i]] = i;
}
int n;
int main() {
	read(n); char ch;
	for(int i = 1; i <= n; ++i) {
        while(!isalpha(ch=getchar()));
        s[i] = ch-'A'+1;
	}
	for(int i = 1; i <= n; ++i) s[(n<<1)-i+1] = s[i];
	Get_Sa(n<<1, 26);
	int l = 1, r = n, tot = 0;
	while(l <= r)
        if(rk[l] < rk[(n<<1)-r+1]) { putchar(s[l++]+'A'-1); if(++tot == 80) tot = 0, putchar('
'); }
        else { putchar(s[r--]+'A'-1); if(++tot == 80) tot = 0, putchar('
'); }
}

CODE(Hash)

从学长那里搬运来的代码(跑得比后缀数组快)
$O(nlogn)$

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#define ull unsigned int
using namespace std;
const int maxn=30003;
char s[maxn];
ull pre[maxn],jc[maxn],val1,val2,pre1[maxn];
short i,j,k,n,m,l,r,mid,len,L,R;
 
inline short getlen(){
    if(s[L]!=s[R])return 0;
    l=1;r=R-L+1;
    if(pre[L+r-1]-pre[L-1]*jc[r]==pre1[R-r+1]-pre1[R+1]*jc[r])return r;r--;
    while(l<r){
        mid=(l+r+1)>>1;
        val1=pre[L+mid-1]-pre[L-1]*jc[mid];
        val2=pre1[R-mid+1]-pre1[R+1]*jc[mid];
        if(val1!=val2)r=mid-1;else l=mid;
        if(s[L+l]!=s[R-l])return l;
    }
    return l;
}
inline bool bigger(){
    if(s[L]!=s[R])return s[L]>s[R];
    len=getlen();
    if(len==R-L+1)return 1;
    else return s[L+len]>s[R-len];
}
int main(){
    scanf("%d",&n);
    for(i=1;i<=n;i++)for(s[i]=getchar();s[i]<'A'||s[i]>'Z';s[i]=getchar());
    jc[0]=1;for(i=1;i<=n;i++)jc[i]=jc[i-1]*107;
    for(i=1;i<=n;i++)pre[i]=pre[i-1]*107+(ull)s[i]-'A';
    for(i=n;i;i--)pre1[i]=pre1[i+1]*107+(ull)s[i]-'A';
    L=1;R=n;
    for(i=1;i<=n;i++)if(bigger()){
        putchar(s[R--]);
        if(i%80==0)putchar('
');
    }else {putchar(s[L++]);if(i%80==0)putchar('
');}
    return 0;
}