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
1 #include <set> 2 #include <map> 3 #include <stack> 4 #include <queue> 5 #include <cmath> 6 #include <ctime> 7 #include <cstdio> 8 #include <string> 9 #include <vector> 10 #include <cstring> 11 #include <iostream> 12 #include <algorithm> 13 #include <unordered_map> 14 15 #define pi acos(-1.0) 16 #define eps 1e-9 17 #define fi first 18 #define se second 19 #define rtl rt<<1 20 #define rtr rt<<1|1 21 #define bug printf("****** ") 22 #define mem(a, b) memset(a,b,sizeof(a)) 23 #define name2str(x) #x 24 #define fuck(x) cout<<#x" = "<<x<<endl 25 #define sfi(a) scanf("%d", &a) 26 #define sffi(a, b) scanf("%d %d", &a, &b) 27 #define sfffi(a, b, c) scanf("%d %d %d", &a, &b, &c) 28 #define sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d) 29 #define sfL(a) scanf("%lld", &a) 30 #define sffL(a, b) scanf("%lld %lld", &a, &b) 31 #define sfffL(a, b, c) scanf("%lld %lld %lld", &a, &b, &c) 32 #define sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d) 33 #define sfs(a) scanf("%s", a) 34 #define sffs(a, b) scanf("%s %s", a, b) 35 #define sfffs(a, b, c) scanf("%s %s %s", a, b, c) 36 #define sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d) 37 #define FIN freopen("../in.txt","r",stdin) 38 #define gcd(a, b) __gcd(a,b) 39 #define lowbit(x) x&-x 40 #define IO iOS::sync_with_stdio(false) 41 42 43 using namespace std; 44 typedef long long LL; 45 typedef unsigned long long ULL; 46 const ULL seed = 13331; 47 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL; 48 const int maxm = 8e6 + 10; 49 const int INF = 0x3f3f3f3f; 50 const int mod = 1e9 + 7; 51 const int maxn = 2e5 + 7; 52 53 struct Suffix_Automaton { 54 int last, tot, nxt[maxn << 1][26], fail[maxn << 1]; 55 int len[maxn << 1];// 最长子串的长度 (该节点子串数量 = len[x] - len[fa[x]]) 56 int sa[maxn << 1], c[maxn << 1]; 57 int sz[maxn << 1];// 被后缀链接的个数，方便求节点字符串的个数 58 LL num[maxn << 1];// 该状态子串的数量 59 LL maxx[maxn << 1];// 长度为x的子串出现次数最多的子串的数目 60 LL sum[maxn << 1];// 该节点后面所形成的自字符串的总数 61 LL subnum, sublen;// subnum表示不同字符串数目,sublen表示不同字符串总长度 62 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 63 void init(int n) { 64 tot = last = 1; 65 fail[1] = len[1] = 0; 66 for (int i = 0; i <= 25; i++) nxt[1][i] = 0; 67 } 68 69 void extend(int c) { 70 int u = ++tot, v = last; 71 for (int i = 0; i <= 25; i++) nxt[u][i] = 0; 72 fail[u] = 0; 73 len[u] = len[v] + 1; 74 num[u] = 1; 75 for (; v && !nxt[v][c]; v = fail[v]) nxt[v][c] = u; 76 if (!v) fail[u] = 1, sz[1]++; 77 else if (len[nxt[v][c]] == len[v] + 1) fail[u] = nxt[v][c], sz[nxt[v][c]]++; 78 else { 79 int now = ++tot, cur = nxt[v][c]; 80 len[now] = len[v] + 1; 81 memcpy(nxt[now], nxt[cur], sizeof(nxt[cur])); 82 fail[now] = fail[cur]; 83 fail[cur] = fail[u] = now; 84 for (; v && nxt[v][c] == cur; v = fail[v]) nxt[v][c] = now; 85 sz[now] += 2; 86 } 87 last = u; 88 } 89 90 void get_posnum() {// 每个节点子串出现的次数 91 for (int i = 1; i <= tot; i++) X[len[i]]++; 92 for (int i = 1; i <= tot; i++) X[i] += X[i - 1]; 93 for (int i = 1; i <= tot; i++) Y[X[len[i]]--] = i; 94 for (int i = tot; i >= 1; i--) num[fail[Y[i]]] += num[Y[i]]; 95 } 96 97 void get_maxx(int n) {// 长度为x的子串出现次数最多的子串的数目 98 get_posnum(); 99 for (int i = 1; i <= tot; i++) maxx[len[i]] = max(maxx[len[i]], num[i]); 100 for (int i = n - 1; i >= 1; i--) maxx[i] = max(maxx[i], maxx[i + 1]); 101 } 102 103 void get_sum() {// 该节点后面所形成的自字符串的总数 104 get_posnum(); 105 for (int i = tot; i >= 1; i--) { 106 sum[Y[i]] = 1; 107 for (int j = 0; j <= 25; j++) sum[Y[i]] += sum[nxt[Y[i]][j]]; 108 } 109 } 110 111 void get_subnum() {//本质不同的子串的个数 112 subnum = 0; 113 for (int i = 1; i <= tot; i++) subnum += len[i] - len[fail[i]]; 114 } 115 116 void get_sublen() {//本质不同的子串的总长度 117 sublen = 0; 118 for (int i = 1; i <= tot; i++) sublen += 1LL * (len[i] + len[fail[i]] + 1) * (len[i] - len[fail[i]]) / 2; 119 } 120 121 void get_sa() { //获取sa数组 122 for (int i = 1; i <= tot; i++) c[len[i]]++; 123 for (int i = 1; i <= tot; i++) c[i] += c[i - 1]; 124 for (int i = tot; i >= 1; i--) sa[c[len[i]]--] = i; 125 } 126 127 int mx[maxn << 1]; 128 129 void match(char s[]) { 130 int p = 1, maxlen = 0; 131 mem(mx, 0); 132 for (int i = 0; s[i]; i++) { 133 int c = s[i] - 'a'; 134 if (nxt[p][c]) p = nxt[p][c], maxlen++; 135 else { 136 for (; p && !nxt[p][c]; p = fail[p]); 137 if (!p) p = 1, maxlen = 0; 138 else maxlen = len[p] + 1, p = nxt[p][c]; 139 } 140 mx[p] = max(mx[p], maxlen); 141 } 142 } 143 144 void get_kth(int k) { 145 int pos = 1, cnt; 146 string s = ""; 147 while (k) { 148 for (int i = 0; i <= 25; i++) { 149 if (nxt[pos][i] && k) { 150 cnt = nxt[pos][i]; 151 if (sum[cnt] < k) k -= sum[cnt]; 152 else { 153 k--; 154 pos = cnt; 155 s += (char) (i + 'a'); 156 break; 157 } 158 } 159 } 160 } 161 cout << s << endl; 162 } 163 164 } sam; 165 166 167 int T, n, cas = 1; 168 char s[maxn], t[maxn]; 169 170 int main() { 171 // FIN; 172 sfi(T); 173 while (T--) { 174 sfi(n); 175 sfs(s + 1); 176 int len = strlen(s + 1); 177 sam.init(len); 178 for (int i = 0; i < n; i++) { 179 sfs(t + 1); 180 len = strlen(t + 1); 181 sam.last = 1; 182 for (int j = 1; j <= len; j++) sam.extend((t[j] - 'a')); 183 } 184 sam.get_subnum(); 185 LL ans1 = sam.subnum; 186 sam.last = 1; 187 len = strlen(s + 1); 188 for (int i = 1; i <= len; i++) sam.extend((s[i] - 'a')); 189 sam.get_subnum(); 190 LL ans2 = sam.subnum; 191 // fuck(ans1),fuck(ans2); 192 printf("Case %d: %lld ", cas++, ans2 - ans1); 193 } 194 return 0; 195 }