bzoj 1692: [Usaco2007 Dec]队列变换

把序列（串）反过来，连到原串上，然后求一下后缀数组，答案就是每次比较a1,a2(a1初始为1，a2初始为n+1),输出rank小的一个

#include<bits/stdc++.h>
#define N 100005<<1
#define LL long long
#define inf 0x3f3f3f3f
using namespace std;
inline int ra()
{
    int x=0,f=1; 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;
}
int p,q,k,n,a[N],v[N],rank[2][N],sa[2][N];
void cal_sa(int sa[N], int rank[N], int Sa[N], int Rank[N])
{
    for (int i=1; i<=n; i++) v[rank[sa[i]]]=i;
    for (int i=n; i>=1; i--)
        if (sa[i]>k)
            Sa[v[rank[sa[i]-k]]--]=sa[i]-k;
    for (int i=n-k+1; i<=n; i++)
        Sa[v[rank[i]]--]=i;
    for (int i=1; i<=n; i++)
        Rank[Sa[i]]=Rank[Sa[i-1]]+(rank[Sa[i-1]]!=rank[Sa[i]] || rank[Sa[i-1]+k]!=rank[Sa[i]+k]);
}
void work()
{
    p=0,q=1; 
    for (int i=1; i<=n; i++) v[a[i]]++;
    for (int i=1; i<=30; i++) v[i]+=v[i-1];
    for (int i=1; i<=n; i++) sa[p][v[a[i]]--]=i;
    for (int i=1; i<=n; i++) rank[p][sa[p][i]]=rank[p][sa[p][i-1]]+(a[sa[p][i]]!=a[sa[p][i-1]]);
    for (k=1; k<n; k<<=1,q^=1,p^=1) cal_sa(sa[p],rank[p],sa[q],rank[q]);
}
int main()
{
    n=ra();
    for (int i=1; i<=n; i++)
    {
        char ch[2]; scanf("%s",ch);
        a[i]=ch[0]-'A'+1;
        a[n-i+1+n]=a[i];
    }
    n<<=1;
    work();
    int a1=1,a2=n/2+1;
    for (int i=1; i<=n>>1; i++)
    {
        if (rank[p][a1]<rank[p][a2]) printf("%c",(char)a[a1++]+'A'-1);
            else printf("%c",(char)a[a2++]+'A'-1);
        if (!(i%80)) cout<<endl;
    }
    return 0;
}