题目链接
题意：给你一个字符串，每次从首或尾取一个字符组成字符串，问所有能够组成的字符串中字典序最小的一个。
思路：首先定义两个指针L和R,如果s[L]<s[R],肯定选s[R],s[L]>s[R]，选s[R].对于s[L]==s[R]的情况,可以考虑L位置的rank以及R位置反串的rank值大小。所以还需要将s的反串构造进去，建立后缀数组。
突然发现其实就一种，因为如果s[L]<s[R]，那么rank[l]一定也小于反串的r的rank值。
#include <iostream>
#include <stdio.h>
#include <cstring>
#include <algorithm>
using namespace std;
#define Max_N 1000100
#define F(x) ((x)/3+((x)%3==1?0:tb))
#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)
int wa1[Max_N*3], wb1[Max_N*3], wv1[Max_N*3], ws1[Max_N*3];
int c0(int *r, int a, int b)
{
return r[a] == r[b] && r[a+1] == r[b+1] && r[a+2] == r[b+2];
}
int c12(int k, int *r, int a, int b)
{
if(k == 2) return r[a] < r[b] || r[a] == r[b] && c12(1, r, a+1, b+1);
else return r[a] < r[b] || r[a] == r[b] && wv1[a+1] < wv1[b+1];
}
void sort(int *r, int *a, int *b, int n, int m)
{
int i;
for (i = 0; i < n; i++) wv1[i] = r[a[i]];
for (i = 0; i < m; i++) ws1[i] = 0;
for (i = 0; i < n; i++) ws1[wv1[i]]++;
for (i = 1; i < m; i++) ws1[i] += ws1[i-1];
for (i = n-1; i >= 0; i--) b[--ws1[wv1[i]]] = a[i];
return;
}
void dc3(int *r, int *sa, int n, int m)
{
int i, j, *rn = r + n, *san = sa + n,ta = 0,tb = (n + 1) / 3,tbc = 0,p;
r[n] = r[n+1] = 0;
for (i = 0; i < n; i++) if (i % 3 != 0) wa1[tbc++] = i;
sort(r+2, wa1, wb1, tbc, m);
sort(r+1, wb1, wa1, tbc, m);
sort(r, wa1, wb1, tbc, m);
for (p = 1,rn[F(wb1[0])] = 0,i = 1;i < tbc; i++)
rn[F(wb1[i])] = c0(r, wb1[i-1], wb1[i]) ? p - 1 : p++;
if (p < tbc) dc3(rn, san, tbc, p);
else for (i = 0; i < tbc; i++) san[rn[i]] = i;
for (i = 0; i < tbc; i++) if (san[i] < tb) wb1[ta++] = san[i] * 3;
if (n % 3 == 1) wb1[ta++] = n - 1;
sort(r, wb1, wa1, ta, m);
for (i = 0; i < tbc; i++) wv1[wb1[i] = G(san[i])] = i;
for (i = 0, j = 0, p = 0; i < ta && j < tbc; p++)
sa[p] = c12(wb1[j] % 3, r, wa1[i], wb1[j]) ? wa1[i++] : wb1[j++];
for(; i < ta; p++) sa[p] = wa1[i++];
for(; j < tbc; p++) sa[p] = wb1[j++];
return;
}
// str sa 都要开三倍的；
void da(int str[], int sa[], int rank1[], int height1[], int n, int m)
{
for (int i = n; i < n*3; i++)
str[i] = 0;
dc3(str, sa, n+1, m);
int i, j, k = 0;
for (i = 0; i <= n; i++) rank1[sa[i]] = i;
for (i = 0; i < n; i++)
{
if (k) k--;
j = sa[rank1[i] - 1];
while (str[i+k] == str[j+k]) k++;
height1[rank1[i]] = k;
}
return;
}
char s[Max_N];
int str[3*Max_N];
int sa[3*Max_N];//sa[i]表示将所有后缀排序后第i小的后缀的编号。
int rank1[3*Max_N];//rank1[i]表示后缀i的排名。
int height1[3*Max_N];
char ans[Max_N];
int main()
{
/*char buf[100];
cin >> buf;
int n = strlen(buf);
//cout << n << endl;
int m = 128;
for (int i = 0; i < strlen(buf); i++)
str[i] = int(buf[i]);
str[n] = 0;
da(str, sa, rank1, height1, n, m);
for (int i = 0; i <= n; i++)
cout << sa[i] << endl;
return 0;
*/
int n;
scanf("%d",&n);
int k=0;
for(int i=0;i<n;i++)
{
char a[2];
scanf("%s",a);
s[k++]=a[0];
}
s[k]='#';
for(int i=0;i<k;i++)
{
s[k+i+1]=s[k-i-1];
}
k=2*n+1;
for(int i=0;i<k;i++)
{
str[i]=int(s[i]);
}
str[k]=0;
int m=128;
da(str, sa, rank1, height1, k, m);
int L=0,R=n-1;
int cnt=0;
while(L<=R)
{
if(s[L]<s[R])
{
ans[cnt++]=s[L];
L++;
}
else if(s[L]>s[R])
{
ans[cnt++]=s[R];
R--;
}
else
{
int m=(n-R);
if(rank1[L]>rank1[n+m])
{
ans[cnt++]=s[R];
R--;
}
else
{
ans[cnt++]=s[L];
L++;
}
}
}
ans[cnt]=' ';
int tot=0;
for(int i=0;i<cnt;i++)
{
printf("%c",ans[i]);
tot++;
if(tot%80==0)
{
printf("
");
}
}
}