后缀数组（仅模板，具体准备区域赛打完学）

后缀数组指的是讲某个字符串的所有后缀按字典序排序后得到的数组。

我们用sa[i]表示在字符串里排行第i个的字符串（按升序排序，及从小到大）是从第sa[i]个开始。

用rk[i]表示第i位开始的后缀在所有后缀里面排第几。

用lcp[i]表示从sa[i]开始的后缀和从sa[i + 1]开始的后缀的最长公共前缀。

有了以上三个数组我门就可以来处理好多好多的东西。

附上模板:

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 80000 + 5;

int wa[N],wb[N],ws[N],wv[N];
bool cmp(int *r, int a, int b, int l) {
    return r[a]==r[b] && r[a+l]==r[b+l];
}
void da(int *r, int *sa, int n, int m){
    int *x = wa, *y = wb;
    for(int i = 0; i < m; i++) ws[i] = 0;
    for(int i = 0; i < n; i++) ws[x[i]=r[i]]++;
    for(int i = 1; i < m; i++) ws[i] += ws[i-1];
    for(int i = n - 1; i >= 0; i--) sa[--ws[x[i]]] = i;
    for(int j = 1, p = 1;p < n; j <<= 1, m = p) {
        p = 0;
        for(int i = n - j; i < n; i++) y[p++] = i;
        for(int i = 0; i < n; i++) if(sa[i] >= j) y[p++] = sa[i]-j;
        for(int i = 0; i < n; i++) wv[i] = x[y[i]];
        for(int i = 0; i < m; i++) ws[i] = 0;
        for(int i = 0; i < n; i++) ws[wv[i]]++;
        for(int i = 1; i < m; i++) ws[i] += ws[i-1];
        for(int i = n - 1; i >= 0; i--) sa[--ws[wv[i]]] = y[i];
        int *t = x; x = y; y = t;
        p = 1; x[sa[0]] = 0;
        for(int i = 1; i < n; i++)
            x[sa[i]] = cmp(y,sa[i-1],sa[i],j) ? p-1 : p++;
    }
}
void getlcp(int *r, int *sa, int *rk, int *lcp, int n) {
    for(int i = 1;i <= n; i++) rk[sa[i]] = i;
    int h = 0;
    lcp[0] = 0;
    lcp[n] = -1;
    for(int i = 0; i < n; i ++){
        int j = sa[rk[i] - 1];
        if(h > 0) h -= 1;
        for(; j + h < n && i + h < n; h ++)
            if(r[i + h] != r[j + h]) break;
        lcp[rk[i] - 1] = h;
    }
}

int n, a[N], rk[N], sa[N], lcp[N];

char s[N];

void solve() {
    int tot = 0;
    int n = strlen(s);
    for(int i = 0; i < n; i++) a[tot++] = s[i]-'0' + 1;
    a[tot ++] = 0;
    da(a, sa, tot, 128);
    getlcp(a, sa, rk, lcp, tot-1);
    for(int i = 0; i < tot; i ++) printf("sa[%d] = %d\n", i, sa[i]);
    for(int i = 0; i < tot; i ++) printf("rk[%d] = %d\n", i, rk[i]);
    for(int i = 0; i < tot; i ++) printf("lcp[%d] = %d\n", i, lcp[i]);
}

int main(){
    while(scanf("%s", s) == 1)solve();
    return 0;
}

　　附上一题：hdu5442

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 100000 + 5;

char s[N], s1[N], s2[N];

int sa[N], rk[N], lcp[N], tmp[N], c[N], t[N], t2[N];

int mn[N][20];

int n, m, le;

void build_sa(char *s, int *sa, int l, int m){
    int *x = t, *y = t2;
    //基数排序
    for(int i = 0; i < m; i ++) c[i] = 0;
    for(int i = 0; i < l; i ++) c[x[i] = s[i]] += 1;
    for(int i = 1; i < m; i ++) c[i] += c[i - 1];
    for(int i = l - 1; i >= 0; i --) sa[--c[x[i]]] = i;
    for(int k = 1; k <= l; k <<= 1){
        int p = 0;
        //直接利用sa数组排序第二关键字
        for(int i = l - k; i < l; i ++) y[p ++] = i;
        for(int i = 0; i < l; i ++) if(sa[i] >= k) y[p ++] = sa[i] - k;
        //基数排序第一关键字
        for(int i = 0; i < m; i ++) c[i] = 0;
        for(int i = 0; i < l; i ++) c[x[y[i]]] += 1;
        for(int i = 1; i < m; i ++) c[i] += c[i - 1];
        for(int i = l - 1; i >= 0; i --) sa[--c[x[y[i]]]] = y[i];
        //根据sa和y数组计算新的x数组
        swap(x, y);
        p = 1;
        x[sa[0]] = 0;
        for(int i = 1; i < l; i ++)
            x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+k] == y[sa[i]+k] ? p - 1 : p ++;
        if(p >= l) break;
        m = p;
    }
}

void build_lcp(char *s, int *sa, int *lcp, int l){
    for(int i = 0; i <= l; i ++) rk[sa[i]] = i;
    int h = 0;
    lcp[0] = 0;
    for(int i = 0; i < l; i ++){
        int j = sa[rk[i] - 1];
        if(h > 0) h -= 1;
        for(;j + h < l && i + h < l; h ++)
            if(s[i + h] != s[j + h]) break;
        lcp[rk[i] - 1] = h;
    }
}

struct RMQ{
    int log2[N];
    void init(int *h, int l){
        for(int i = 0; i <= l; i ++)
            log2[i] = (i == 0 ? -1 : log2[i >> 1] + 1);
        for(int i = 0; i < l; i ++) mn[i][0] = h[i];
        for(int j = 1; j < 20; j ++)
            for(int i = 1; i + (1 << j) <= l + 1; i ++)
                mn[i][j] = min(mn[i][j - 1], mn[i + (1 << j - 1)][j - 1]);
    }
    int query(int ql, int qr){
        int k = log2[qr - ql + 1];
        return min(mn[ql][k], mn[qr - (1 << k) + 1][k]);
    }
}rmq;

int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        scanf("%d", &le);
        scanf("%s", s);
        memset(s1, 0, sizeof(s1));
        memset(s2, 0, sizeof(s2));
        for(int i = 0; i < le; i ++)s[i + le] = s[i];
        s[2 * le] = '\0';
        le *= 2;
        build_sa(s, sa, le + 1, 128);
        int ans1 = sa[le] + 1;
        //printf("ans1 = %d\n", ans1);
        for(int i = 0; i < le / 2; i ++)s1[i] = s[ans1 + i - 1];
        s1[le/2] = '\0';
        reverse(s, s + le);
        build_sa(s, sa, le + 1, 128);
        build_lcp(s, sa, lcp, le);
        /*
        for(int i = 0; i <= le; i ++)printf("sa[%d] = %d\n", i, sa[i]);
        for(int i = 0; i <= le; i ++)printf("rk[%d] = %d\n", i, rk[i]);
        for(int i = 0; i <= le; i ++)printf("lcp[%d] = %d\n", i, lcp[i]);
        */
        int ans2 = sa[le];
        int st = le;
        //printf("le = %d\n", le);
        for(int i = le - 1; i >= 0; i --){
            st =min(st, lcp[i]);
            //printf("sa[%d] = %d\n", i, sa[i]);
            if(st < le / 2) break;
            if(sa[i] < le / 2)ans2 = max(sa[i], ans2);
        }
        for(int i = 0; i < le / 2; i ++)s2[i] = s[ans2 + i];
        ans2 = le/2 - ans2;
        s2[le/2] = '\0';
        //puts(s1), puts(s2);
        int flag = strcmp(s1, s2);
        if(flag > 0) printf("%d 0\n", ans1);
        if(flag < 0) printf("%d 1\n", ans2);
        if(!flag){
            if(ans1 <= ans2) printf("%d 0\n", ans1);
            else printf("%d 1\n", ans2);
        }
    }
    return 0;
}

后缀数组还可以处理最长公共子串：传送门。　

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int INF = 0x7fffffff;

const int N = 200000 + 5;

int wa[N],wb[N],ws[N],wv[N];

int n, a[N], rk[N], sa[N], lcp[N];

char s[N], t[N];

bool cmp(int *r, int a, int b, int l) {
    return r[a]==r[b] && r[a+l]==r[b+l];
}
void da(int *r, int *sa, int n, int m){
    int *x = wa, *y = wb;
    for(int i = 0; i < m; i++) ws[i] = 0;
    for(int i = 0; i < n; i++) ws[x[i]=r[i]]++;
    for(int i = 1; i < m; i++) ws[i] += ws[i-1];
    for(int i = n - 1; i >= 0; i--) sa[--ws[x[i]]] = i;
    for(int j = 1, p = 1;p < n; j <<= 1, m = p) {
        p = 0;
        for(int i = n - j; i < n; i++) y[p++] = i;
        for(int i = 0; i < n; i++) if(sa[i] >= j) y[p++] = sa[i]-j;
        for(int i = 0; i < n; i++) wv[i] = x[y[i]];
        for(int i = 0; i < m; i++) ws[i] = 0;
        for(int i = 0; i < n; i++) ws[wv[i]]++;
        for(int i = 1; i < m; i++) ws[i] += ws[i-1];
        for(int i = n - 1; i >= 0; i--) sa[--ws[wv[i]]] = y[i];
        int *t = x; x = y; y = t;
        p = 1; x[sa[0]] = 0;
        for(int i = 1; i < n; i++)
            x[sa[i]] = cmp(y,sa[i-1],sa[i],j) ? p-1 : p++;
    }
}
void getlcp(int *r, int *sa, int *rk, int *lcp, int n) {
    for(int i = 1;i <= n; i++) rk[sa[i]] = i;
    int h = 0;
    lcp[0] = 0;
    lcp[n] = -1;
    for(int i = 0; i < n; i ++){
        int j = sa[rk[i] - 1];
        if(h > 0) h -= 1;
        for(; j + h < n && i + h < n; h ++)
            if(r[i + h] != r[j + h]) break;
        lcp[rk[i] - 1] = h;
    }
}

int main(){
    while(scanf("%s%s", s, t) == 2){
        int ls = strlen(s);
        int lt = strlen(t);
        int l = 0;
        for(int i = 0; i < ls; i ++)a[l++] = s[i] - 'a' + 1;
        a[l++] = 27;
        for(int i = 0; i < lt; i ++)a[l++] = t[i] - 'a' + 1;
        a[l++] = 0;
        da(a, sa, l, 32);
        getlcp(a, sa, rk, lcp, l - 1);
        int ans = 0;
        for(int i = 0; i < l; i ++)
            if((sa[i] < ls) != (sa[i + 1] < ls))ans = max(ans, lcp[i]);
        printf("%d\n", ans);
    }
    return 0;
}