1427 - Substring Frequency (II)
InputA string is a finite sequence of symbols that are chosen from an alphabet. In this problem you are given a string T and n queries each with a string Pi, where the strings contain lower case English alphabets only. You have to find the number of times Pi occurs as a substring of T.
Input starts with an integer T (≤ 10), denoting the number of test cases.
Each case starts with a line containing an integer n (1 ≤ n ≤ 500). The next line contains the string T (1 ≤ |T| ≤ 106). Each of the next n lines contains a string Pi (1 ≤ |Pi| ≤ 500).
Output
For each case, print the case number in a single line first. Then for each string Pi, report the number of times it occurs as a substring of T in a single line.
Sample Input |
Output for Sample Input |
25ababacbabcababa
ac a abc 3 lightoj oj light lit |
Case 1:2314
1 Case 2: 1 1 0 |
Notes
- Dataset is huge, use faster I/O methods.
- If S is a string then |S| denotes the length of S.
题目类型:AC自动机
算法分析:将子串插入到Trie树中,然后使用cnt数组统计在文本串T中,相应子串出现的个数即可,注意子串可能出现重复的情况!!!由于n比较小,所以可以使用O(n^2)的枚举将标号后面重复的相应子串的个数求解得到
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 |
#include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> #define lson rt << 1, l, m #define rson rt << 1 | 1, m + 1, r using namespace std; const int INF = 0x7FFFFFFF; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int MOD = 1e9 + 7; const int maxn = 26; const int nodemaxn = 250000 + 66; char ss[1000000+maxn]; int cnt[500+maxn]; char val[500+maxn][500+maxn]; struct ACNode { int next[nodemaxn][maxn], pre[nodemaxn], last[nodemaxn], id[nodemaxn]; int root, len; void Init () { len = 0; root = newnode (); } int newnode () { for (int i = 0; i < maxn; i++) next[len][i] = -1; last[len] = id[len] = 0; len++; return len - 1; } void Insert (char *s, int ff) { int p = root; for(int i = 0; s[i]; i++) { if(next[p][s[i]-'a'] == -1) next[p][s[i]-'a'] = newnode (); p = next[p][s[i]-'a']; } last[p]++; id[p] = ff; } void Build () { queue <int> qu; pre[root] = root; for(int i = 0; i < maxn; i++) { if(next[root][i] == -1) next[root][i] = root; else { pre[next[root][i]] = root; qu.push (next[root][i]); } } while (!qu.empty ()) { int p = qu.front(); qu.pop (); for (int i = 0; i < maxn; i++) { if (next[p][i] == -1) next[p][i] = next[pre[p]][i]; else { pre[next[p][i]] = next[pre[p]][i]; qu.push (next[p][i]); } } } } void Query (char *s) { int p = root; for (int i = 0; s[i]; i++) { p = next[p][s[i]-'a'];//一步到位,重要!!! 当存在时走到下一层,否则回溯!!! int temp = p; while (temp != root) { if (last[temp]) { cnt[id[temp]]++; } temp = pre[temp]; } } } }; ACNode ac; int main() { // freopen ("aaa.txt", "r", stdin); int n, t, flag = 1; scanf ("%d", &t); while (t--) { printf ("Case %d:\n", flag++); scanf ("%d", &n); scanf ("%s", ss); ac.Init (); memset (cnt, 0, sizeof (cnt)); for (int i = 1; i <= n; i++) { scanf ("%s", val[i]); int j; for (j = 1; j < i; j++) if (!strcmp (val[j], val[i])) break; if (j >= i) ac.Insert (val[i], i); } ac.Build (); ac.Query (ss); for (int i = 1; i <= n; i++) { int j; for (j = 1; j < i; j++) if (!strcmp (val[j], val[i])) break; if (j < i) printf ("%d\n", cnt[j]); else printf ("%d\n", cnt[i]); } } return 0; } |
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