不在B中的A的子串数量 HDU - 4416 (后缀自动机模板题目)

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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

 

技术图片
  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("******\n")
 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\n", cas++, ans2 - ans1);
193     
194     return 0;
195 
View Code

 

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