后缀数组模板详解。

给定一个字符串S，比如它是（abcad）,那么它的后缀有”abcad“  ,"bcad ", "cad",  "ad", "d", ""。 讲这些后缀字符串按照字典序排序，得到的就是后缀数组。如果用普通的排序方法，排序要o（nlogn），但是每两个字符比较大小要O（n），所以是o（n × n log n）的复杂度。但是利用特殊的算法可以将其降到o（nlogn）。 

该算法的思想是，先将每个后缀数组按照只考虑头一个字符，排序， 再将其按照只考虑头两个字符排序，然后是头4个，然后是8个，直到考虑n个。

#include <iostream>
#include <stdio.h>
#include <cstring>
#include <algorithm>
using namespace std;
#define Max_N (200000 + 100)
int n;
int k;

int rank1[Max_N], tmp[Max_N];
int sa[Max_N];// n为字符长度， rank1[[i] 表示的是以第i个字符开头的后缀数组的排名， sa[i] 是排在i位的后缀数组的开头字符的位置。
bool compare_sa(int i, int j)
{
	if(rank1[i] != rank1[j]) return rank1[i] < rank1[j];
	else {
		int ri = i + k <= n ? rank1[i + k] : -1;
		int rj = j + k <= n ? rank1[j + k] : -1;
		return ri < rj;
	}
}

void construct_sa(string buf)
{
	n = buf.length();
	for (int i = 0; i <= n; i++) {
		sa[i] = i;
		rank1[i] = i < n ? buf[i] : -1;
	}

	for ( k = 1; k <= n; k *= 2) {
		sort(sa, sa + n +1, compare_sa);

		tmp[sa[0]] = 0;
		for (int i = 1; i <= n; i++) {
			tmp[sa[i]] = tmp[sa[i-1]] + (compare_sa(sa[i-1], sa[i]) ? 1 : 0);
		}
		for (int i = 0; i <= n; i++) {
			rank1[i] = tmp[i];
		}
	}
}
int main()
{ 
	string str;
	cin >> str;
	construct_sa(str);
	for (int i = 0; i <= n; i++) {
		cout << sa[i] << endl;
	}
	cout << endl;
	for (int i = 0; i <= n; i++) {
		cout << rank1[i] << endl;
	}
	return 0;	
}

cd3算法，没怎么理解， 所以只能照敲了模板，保存一发。

cd3的算法虽然是线性的o(n) 的，但是其前面的常数特别大而且会占用大量内存，相对于倍增算法的o(nlogn)优势也不是很大，只要题目不是特别特殊，倍增算法都可以解决。

#include <iostream>
#include <stdio.h>
#include <cstring>
#include <algorithm>
using namespace std;
#define Max_N 1000
#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;
}
int str[3*Max_N];
int sa[3*Max_N];
int rank1[3*Max_N];
int height1[3*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;
		

}