不在B中的A的子串数量 HDU

题目：

给定一个字符串a，又给定一系列b字符串，求字符串a的子串不在b中出现的个数。

题解：

先将所有的查询串放入后缀自动机(每次将sam.last=1)（算出所有子串个数）

然后将母串放入后缀自动机然后记录这个子串个数

两个值相减即可

#include <set>
#include <map>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>

#define  pi    acos(-1.0)
#define  eps   1e-9
#define  fi    first
#define  se    second
#define  rtl   rt<<1
#define  rtr   rt<<1|1
#define  bug                printf("******
")
#define  mem(a, b)          memset(a,b,sizeof(a))
#define  name2str(x)        #x
#define  fuck(x)            cout<<#x" = "<<x<<endl
#define  sfi(a)             scanf("%d", &a)
#define  sffi(a, b)         scanf("%d %d", &a, &b)
#define  sfffi(a, b, c)     scanf("%d %d %d", &a, &b, &c)
#define  sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
#define  sfL(a)             scanf("%lld", &a)
#define  sffL(a, b)         scanf("%lld %lld", &a, &b)
#define  sfffL(a, b, c)     scanf("%lld %lld %lld", &a, &b, &c)
#define  sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d)
#define  sfs(a)             scanf("%s", a)
#define  sffs(a, b)         scanf("%s %s", a, b)
#define  sfffs(a, b, c)     scanf("%s %s %s", a, b, c)
#define  sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d)
#define  FIN                freopen("../in.txt","r",stdin)
#define  gcd(a, b)          __gcd(a,b)
#define  lowbit(x)          x&-x
#define  IO                 iOS::sync_with_stdio(false)


using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const ULL seed = 13331;
const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
const int maxm = 8e6 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 7;

struct Suffix_Automaton {
    int last, tot, nxt[maxn << 1][26], fail[maxn << 1];
    int len[maxn << 1];// 最长子串的长度 (该节点子串数量 = len[x] - len[fa[x]])
    int sa[maxn << 1], c[maxn << 1];
    int sz[maxn << 1];// 被后缀链接的个数，方便求节点字符串的个数
    LL num[maxn << 1];// 该状态子串的数量
    LL maxx[maxn << 1];// 长度为x的子串出现次数最多的子串的数目
    LL sum[maxn << 1];// 该节点后面所形成的自字符串的总数
    LL subnum, sublen;// subnum表示不同字符串数目,sublen表示不同字符串总长度
    int X[maxn << 1], Y[maxn << 1]; // Y表示排名为x的节点，X表示该长度前面还有多少个
    void init(int n) {
        tot = last = 1;
        fail[1] = len[1] = 0;
        for (int i = 0; i <= 25; i++) nxt[1][i] = 0;
    }

    void extend(int c) {
        int u = ++tot, v = last;
        for (int i = 0; i <= 25; i++) nxt[u][i] = 0;
        fail[u] = 0;
        len[u] = len[v] + 1;
        num[u] = 1;
        for (; v && !nxt[v][c]; v = fail[v]) nxt[v][c] = u;
        if (!v) fail[u] = 1, sz[1]++;
        else if (len[nxt[v][c]] == len[v] + 1) fail[u] = nxt[v][c], sz[nxt[v][c]]++;
        else {
            int now = ++tot, cur = nxt[v][c];
            len[now] = len[v] + 1;
            memcpy(nxt[now], nxt[cur], sizeof(nxt[cur]));
            fail[now] = fail[cur];
            fail[cur] = fail[u] = now;
            for (; v && nxt[v][c] == cur; v = fail[v]) nxt[v][c] = now;
            sz[now] += 2;
        }
        last = u;
    }

    void get_posnum() {// 每个节点子串出现的次数
        for (int i = 1; i <= tot; i++) X[len[i]]++;
        for (int i = 1; i <= tot; i++) X[i] += X[i - 1];
        for (int i = 1; i <= tot; i++) Y[X[len[i]]--] = i;
        for (int i = tot; i >= 1; i--) num[fail[Y[i]]] += num[Y[i]];
    }

    void get_maxx(int n) {// 长度为x的子串出现次数最多的子串的数目
        get_posnum();
        for (int i = 1; i <= tot; i++) maxx[len[i]] = max(maxx[len[i]], num[i]);
        for (int i = n - 1; i >= 1; i--) maxx[i] = max(maxx[i], maxx[i + 1]);
    }

    void get_sum() {// 该节点后面所形成的自字符串的总数
        get_posnum();
        for (int i = tot; i >= 1; i--) {
            sum[Y[i]] = 1;
            for (int j = 0; j <= 25; j++) sum[Y[i]] += sum[nxt[Y[i]][j]];
        }
    }

    void get_subnum() {//本质不同的子串的个数
        subnum = 0;
        for (int i = 1; i <= tot; i++) subnum += len[i] - len[fail[i]];
    }

    void get_sublen() {//本质不同的子串的总长度
        sublen = 0;
        for (int i = 1; i <= tot; i++) sublen += 1LL * (len[i] + len[fail[i]] + 1) * (len[i] - len[fail[i]]) / 2;
    }

    void get_sa() { //获取sa数组
        for (int i = 1; i <= tot; i++) c[len[i]]++;
        for (int i = 1; i <= tot; i++) c[i] += c[i - 1];
        for (int i = tot; i >= 1; i--) sa[c[len[i]]--] = i;
    }

    int mx[maxn << 1];

    void match(char s[]) {
        int p = 1, maxlen = 0;
        mem(mx, 0);
        for (int i = 0; s[i]; i++) {
            int c = s[i] - 'a';
            if (nxt[p][c]) p = nxt[p][c], maxlen++;
            else {
                for (; p && !nxt[p][c]; p = fail[p]);
                if (!p) p = 1, maxlen = 0;
                else maxlen = len[p] + 1, p = nxt[p][c];
            }
            mx[p] = max(mx[p], maxlen);
        }
    }

    void get_kth(int k) {
        int pos = 1, cnt;
        string s = "";
        while (k) {
            for (int i = 0; i <= 25; i++) {
                if (nxt[pos][i] && k) {
                    cnt = nxt[pos][i];
                    if (sum[cnt] < k) k -= sum[cnt];
                    else {
                        k--;
                        pos = cnt;
                        s += (char) (i + 'a');
                        break;
                    }
                }
            }
        }
        cout << s << endl;
    }

} sam;


int T, n, cas = 1;
char s[maxn], t[maxn];

int main() {
    //  FIN;
    sfi(T);
    while (T--) {
        sfi(n);
        sfs(s + 1);
        int len = strlen(s + 1);
        sam.init(len);
        for (int i = 0; i < n; i++) {
            sfs(t + 1);
            len = strlen(t + 1);
            sam.last = 1;
            for (int j = 1; j <= len; j++) sam.extend((t[j] - 'a'));
        }
        sam.get_subnum();
        LL ans1 = sam.subnum;
        sam.last = 1;
        len = strlen(s + 1);
        for (int i = 1; i <= len; i++) sam.extend((s[i] - 'a'));
        sam.get_subnum();
        LL ans2 = sam.subnum;
        // fuck(ans1),fuck(ans2);
        printf("Case %d: %lld
", cas++, ans2 - ans1);
    }
    return 0;
}