JJM

我的头文件

#include <bits/stdc++.h>
#define rep(i,x,y) for(int i=x;i<=y;i++)
#define per(i,x,y) for(int i=x;i>=y;i--)
#define debug(a) cerr<<#a<<"=="<<a<<endl
#define iout(x) printf("%d
",x)
#define lout(x) printf("%lld
",x)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef vector<int> VI;
typedef long long ll;
typedef pair<int, int> PII;
const int mod=1e9+7;
template<typename T> inline void read (T &x) {
    x = 0; T f = 1;
    char ch;
    do {
        ch = getchar(); if (ch == '-') f = -1;
    } while (ch < '0' || ch > '9');
    do x = x * 10 + ch - '0', ch = getchar();
    while (ch <= '9' && ch >= '0'); x *= f;
}

template<typename A, typename B> inline void read (A &x, B &y) {read (x);read (y);}
template<typename A, typename B, typename C> inline void read (A &x, B &y, C &z) {read (x);read (y);read (z);}
template<typename A, typename B, typename C, typename D> inline void read (A &x, B &y, C &z, D &w) {read (x);read (y);read (z);read (w);}
template<typename A,typename B> inline A fexp(A x,B p){A ans=1;for(;p;p>>=1,x=1LL*x*x%mod)if(p&1)ans=1LL*ans*x%mod;return ans;}
template<typename A,typename B> inline A fexp(A x,B p,A mo){A ans=1;for(;p;p>>=1,x=1LL*x*x%mo)if(p&1)ans=1LL*ans*x%mo;return ans;}
//head

suffix array

#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
using namespace std;
const int maxn = 200000 + 100;

int sa[maxn];
int t1[maxn], t2[maxn], c[maxn];
int rak[maxn], height[maxn];
int r[maxn];
//待排序的字符串放在s数组中，从r[0]到r[n-1],长度为n,且最大值小于m,
//除r[n-1]外的所有r[i]都大于0，r[n-1]=0
//函数结束以后结果放在sa数组中
void build_sa(int r[], int n, int m) {
    int i, j, p, *x = t1, *y = t2;
    //第一轮基数排序，如果s的最大值很大，可改为快速排序
    for (i = 0; i < m; i++)c[i] = 0;
    for (i = 0; i < n; i++)c[x[i] = r[i]]++;
    for (i = 1; i < m; i++)c[i] += c[i - 1];
    for (i = n - 1; i >= 0; i--)sa[--c[x[i]]] = i;
    for (j = 1; j <= n; j <<= 1) {
        p = 0;
        //直接利用sa数组排序第二关键字
        for (i = n - j; i < n; i++)y[p++] = i; //后面的j个数第二关键字为空的最小
        for (i = 0; i < n; i++)if (sa[i] >= j)y[p++] = sa[i] - j;
        //这样数组y保存的就是按照第二关键字排序的结果
        //基数排序第一关键字
        for (i = 0; i < m; i++)c[i] = 0;
        for (i = 0; i < n; i++)c[x[y[i]]]++;
        for (i = 1; i < m; i++)c[i] += c[i - 1];
        for (i = n - 1; i >= 0; i--)sa[--c[x[y[i]]]] = y[i];
        //根据sa和x数组计算新的x数组
        swap(x, y);
        p = 1; x[sa[0]] = 0;
        for (i = 1; i < n; i++)
            x[sa[i]] = y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + j] == y[sa[i] + j] ? p - 1 : p++;
        if (p >= n)break;
        m = p; //下次基数排序的最大值
    }
}
void getHeight(int r[], int n) {
    int i, j, k = 0;
    for (i = 0; i <= n; i++)rak[sa[i]] = i;
    for (i = 0; i < n; i++) {
        if (k)k--;
        j = sa[rak[i] - 1];
        while (r[i + k] == r[j + k])k++;
        height[rak[i]] = k;
    }
}
char str[maxn];


int main() {
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int T;
    scanf("%d", &T);
    while (T--) {
        scanf("%s", str);
        int n = strlen(str);
        for (int i = 0; i <= n; i++)r[i] = str[i];
        build_sa(r, n + 1, 128);
        getHeight(r, n);
        int ans = n * (n + 1) / 2;
        for (int i = 2; i <= n; i++)ans -= height[i];
        printf("%d
", ans);
    }
    return 0;
}

SAM

#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std;
const int N = 100010;//点的数目
// 空间复杂度不多余2*n
// 下表为0也使用
struct SAM {
    struct Node {
        int ch[26];
        int f, len;
        void init() { f = -1, len = 0; memset(ch, 0xff, sizeof (ch));}
    };
    Node sa[N<<1];
    int idx, last;
    void init() {
        idx = last = 0;
        sa[idx++].init();
    }
    int newnode() {
        sa[idx].init();
        return idx++;
    }
    void add(int c) {
        int end = newnode();
        int tmp = last;
        sa[end].len = sa[last].len + 1;
        for ( ; tmp != -1 && sa[tmp].ch[c] == -1; tmp = sa[tmp].f) {
            sa[tmp].ch[c] = end;
        }
        if (tmp == -1) sa[end].f = 0; // 所有的上一轮可接受点都没有指向字符c的孩子节点
        else {
            int nxt = sa[tmp].ch[c];
            if (sa[tmp].len + 1 == sa[nxt].len) sa[end].f = nxt; // 如果可接受点有向c的转移，且长度只加1，那么该孩子可以替代当前的end，并且end的双亲指向该孩子
            else {
                int np = newnode();
                sa[np] = sa[nxt];
                sa[np].len = sa[tmp].len + 1;
                sa[end].f = sa[nxt].f = np;
                for (; tmp != -1 && sa[tmp].ch[c] == nxt; tmp = sa[tmp].f) {
                    sa[tmp].ch[c] = np;
                }
            }
        }
        last = end;
    }
};
void upd(int& x,int y) {
    if(y>x) x=y;
}
SAM sam;
char s1[N],s2[N];
int main() {
    freopen("in.txt","r",stdin);            
    while(scanf("%s%s",s1,s2)!=EOF) {
        int len1=strlen(s1);
        int len2=strlen(s2);
        sam.init();
        for(int i=0;i<len1;i++) {
            sam.add(s1[i]-'a');
        }
        int p=0,rms=0;
        int ans=0;
        for(int i=0;i<len2;i++) {
            int x=s2[i]-'a';
            if(sam.sa[p].ch[x]!=-1) {
                rms++;
                p=sam.sa[p].ch[x];
            }else {
                for(;p!=-1&&sam.sa[p].ch[x]==-1;p=sam.sa[p].f);
                if(p==-1) rms=0,p=0;
                else {
                    rms=sam.sa[p].len+1;
                    p=sam.sa[p].ch[x];
                }
            }
            upd(ans,rms);
        }
        printf("%d
",ans);
    }
    return 0;
}