Description
很久很久以前，在你刚刚学习字符串匹配的时候，有两个仅包含小写字母的字符串A和B，其中A串长度为m，B串长度为n。可当你现在再次碰到这两个串时，这两个串已经老化了，每个串都有不同程度的残缺。
你想对这两个串重新进行匹配，其中A为模板串，那么现在问题来了，请回答，对于B的每一个位置i，从这个位置开始连续m个字符形成的子串是否可能与A串完全匹配?
Input
第一行包含两个正整数m,n(1<=m<=n<=300000)，分别表示A串和B串的长度。
第二行为一个长度为m的字符串A。
第三行为一个长度为n的字符串B。
两个串均仅由小写字母和*号组成，其中*号表示相应位置已经残缺。
Output
第一行包含一个整数k，表示B串中可以完全匹配A串的位置个数。
若k>0，则第二行输出k个正整数，从小到大依次输出每个可以匹配的开头位置（下标从1开始）。
Sample Input
3 7
a*b
aebr*ob
Sample Output
2
1 5
HINT
Source
Solution
对于每个字母的权值就设为1~26。*的权值设为0。两个字符可以匹配当且仅当A_i∗B_j∗(A_i−B_j )^2等于0。那么我们将这个式子展开，展开以后的每一项分别用FFT计算即可。
1 #include <cstdio>
2 #include <algorithm>
3 #include <cmath>
4 #include <cstring>
5
6 #define R register
7 #define maxn 1048576
8 typedef long double db;
9 const db pi = acosl(-1);
10 char A[maxn], B[maxn];
11 struct Complex {
12 db x, y;
13 inline Complex operator - (const Complex &that) const {return (Complex) {x - that.x, y - that.y};}
14 inline Complex operator * (const Complex &that) const {return (Complex) {x * that.x - y * that.y, x * that.y + y * that.x};}
15 inline void operator += (const Complex &that) {x += that.x; y += that.y;}
16 } w[maxn];
17 int N;
18 void init()
19 {
20 R int h = N >> 1;
21 for (R int i = 0; i < h; ++i) w[i + h] = (Complex) {cos(2 * pi / N * i), sin(2 * pi / N * i)};
22 for (R int i = h; i--; ) w[i] = w[i << 1];
23 }
24 void bit_reverse(R Complex *a, R Complex *b)
25 {
26 for (R int i = 0; i < N; ++i) b[i] = a[i];
27 for (R int i = 0, j = 0; i < N; ++i)
28 {
29 i > j ? std::swap(b[i], b[j]), 1 : 0;
30 for (R int l = N >> 1; (j ^= l) < l; l >>= 1);
31 }
32 }
33 void dft(R Complex *a)
34 {
35 for (R int l = 2, m = 1; m != N; l <<= 1, m <<= 1)
36 for (R int i = 0; i < N; i += l)
37 for (R int j = 0; j < m; ++j)
38 {
39 R Complex tmp = a[i + j + m] * w[j + m];
40 a[i + j + m] = a[i + j] - tmp;
41 a[i + j] += tmp;
42 }
43 }
44 Complex a[maxn], b[maxn], ta[maxn], tb[maxn], tc[maxn], ans[maxn];
45 int aa[maxn], bb[maxn];
46 int main()
47 {
48 R int la, lb;
49 scanf("%s%s", A, B);
50 la = strlen(A); lb = strlen(B);
51 for (N = 1; N < (la + lb); N <<= 1);
52 init();
53 std::reverse(A, A + la);
54 for (R int i = 0; i < la; ++i) A[i] == ‘*‘ ? 0 : aa[i] = A[i] - ‘a‘ + 1;
55 for (R int i = 0; i < lb; ++i) B[i] == ‘*‘ ? 0 : bb[i] = B[i] - ‘a‘ + 1;
56
57 for (R int i = 0; i < la; ++i) a[i].x = aa[i] * aa[i] * aa[i], a[i].y = 0;
58 for (R int i = 0; i < lb; ++i) b[i].x = bb[i], b[i].y = 0;
59 bit_reverse(a, ta); bit_reverse(b, tb);
60 dft(ta); dft(tb);
61 for (R int i = 0; i < N; ++i) ans[i] += (ta[i] * tb[i]);
62
63 for (R int i = 0; i < la; ++i) a[i].x = -2 * aa[i] * aa[i], a[i].y = 0;
64 for (R int i = 0; i < lb; ++i) b[i].x = bb[i] * bb[i], b[i].y = 0;
65 bit_reverse(a, ta); bit_reverse(b, tb);
66 dft(ta); dft(tb);
67 for (R int i = 0; i < N; ++i) ans[i] += (ta[i] * tb[i]);
68
69 for (R int i = 0; i < la; ++i) a[i].x = aa[i], a[i].y = 0;
70 for (R int i = 0; i < lb; ++i) b[i].x = bb[i] * bb[i] * bb[i], b[i].y = 0;
71 bit_reverse(a, ta); bit_reverse(b, tb);
72 dft(ta); dft(tb);
73 for (R int i = 0; i < N; ++i) ans[i] += (ta[i] * tb[i]);
74
75 std::reverse(ans + 1, ans + N);
76 bit_reverse(ans, tc);
77 dft(tc);
78 // for (R int i = 0; i < N; ++i) printf("%lf %lf\n", tc[i].x, tc[i].y);
79 for (R int i = la - 1; i < lb; ++i) if (fabs(tc[i].x / N) < 0.5) printf("%d ", i - la + 2);
80 return 0;
81 }
82 /*
83 399906 399924 399942 399960 399978 399996
84 */
FFT
其实这题的数据用bitset是可以水过的。不过可以卡。然后我优化了一发常数就只是最慢的数据本机3s以内了。我把代码放在这里欢迎大家来hack! O(∩_∩)O~~
1 #include <cstdio>
2 #include <cstring>
3 #include <bitset>
4 #include <algorithm>
5
6 #define R register
7 #define maxn 500010
8 #define maxs 15635
9 //#define inline
10 char A[maxn], B[maxn];
11 /*inline bool eq(R char a, R char b)
12 {
13 return a == b || a == ‘*‘ || b == ‘*‘;
14 }*/
15 typedef unsigned uint;
16 int len;
17
18 #define set(v, x) (v[x >> 5] |= 1u << (x & 31))
19 #define query(v, x) ((1u << (x & 31)) & v[x >> 5])
20 uint p[26][32][maxs];
21 uint aph[26][maxs], ans[maxs];
22 inline void init(R int c)
23 {
24 R uint *t = p[c][0], *tt;
25 for (R int i = 0; i <= len; ++i) t[i] = aph[c][i];
26 for (R int i = 1; i < 32; ++i)
27 {
28 t = p[c][i]; tt = p[c][i - 1];
29 for (R int j = 0; j <= len; ++j)
30 t[j] = ((tt[j] >> 1) | ((tt[j + 1] & 1u) << 31));
31 }
32 }
33 int cnt;
34 inline void bit_and(R int c, R int l)
35 {
36 R int fir = l >> 5; R uint *t = p[c][l & 31];
37 R int i = fir, j = 0;
38 for (; i + 36 < len; i += 36, j += 36)
39 {
40 ans[j] &= t[i];
41 ans[j + 1] &= t[i + 1];
42 ans[j + 2] &= t[i + 2];
43 ans[j + 3] &= t[i + 3];
44 ans[j + 4] &= t[i + 4];
45 ans[j + 5] &= t[i + 5];
46 ans[j + 6] &= t[i + 6];
47 ans[j + 7] &= t[i + 7];
48 ans[j + 8] &= t[i + 8];
49 ans[j + 9] &= t[i + 9];
50 ans[j + 10] &= t[i + 10];
51 ans[j + 11] &= t[i + 11];
52 ans[j + 12] &= t[i + 12];
53 ans[j + 13] &= t[i + 13];
54 ans[j + 14] &= t[i + 14];
55 ans[j + 15] &= t[i + 15];
56 ans[j + 16] &= t[i + 16];
57 ans[j + 17] &= t[i + 17];
58 ans[j + 18] &= t[i + 18];
59 ans[j + 19] &= t[i + 19];
60 ans[j + 20] &= t[i + 20];
61 ans[j + 21] &= t[i + 21];
62 ans[j + 22] &= t[i + 22];
63 ans[j + 23] &= t[i + 23];
64 ans[j + 24] &= t[i + 24];
65 ans[j + 25] &= t[i + 25];
66 ans[j + 26] &= t[i + 26];
67 ans[j + 27] &= t[i + 27];
68 ans[j + 28] &= t[i + 28];
69 ans[j + 29] &= t[i + 29];
70 ans[j + 30] &= t[i + 30];
71 ans[j + 31] &= t[i + 31];
72 ans[j + 32] &= t[i + 32];
73 ans[j + 33] &= t[i + 33];
74 ans[j + 34] &= t[i + 34];
75 ans[j + 35] &= t[i + 35];
76 }
77 for (; i <= len; ++i, ++j) ans[j] &= t[i];
78 }
79
80 //std::bitset<maxn> aph[26], ans;
81 int fail[maxn];
82 int r[maxn];
83 inline bool cmp(R int a, R int b) {return A[a] < A[b] || (A[a] == A[b] && (a & 31) < (b & 31));}
84 int main()
85 {
86 // freopen("str.in", "r", stdin);
87 // freopen("str.out", "w", stdout);
88 scanf("%s%s", A, B);
89 R int la = strlen(A), lb = strlen(B);
90 len = lb >> 5;
91 R int xcnt = 0;
92 for (R int i = 0; i < la; ++i) xcnt += A[i] == ‘*‘;
93 for (R int i = 0; i < lb; ++i) xcnt += B[i] == ‘*‘;
94 /*if (!xcnt)
95 {
96 fail[1] = 0;
97 for (R int i = la; i; --i) A[i] = A[i - 1];
98 for (R int i = lb; i; --i) B[i] = B[i - 1];
99 for (R int i = 2, p = 0; i <= la; ++i)
100 {
101 while (p && A[p + 1] != A[i]) p = fail[p];
102 A[p + 1] == A[i] ? ++p : 0;
103 fail[i] = p;
104 }
105 for (R int i = 1, p = 0; i <= lb; ++i)
106 {
107 while (p && A[p + 1] != B[i]) p = fail[p];
108 A[p + 1] == B[i] ? ++p : 0;
109 if (p == la)
110 {
111 printf("%d ", i - la + 1);
112 p = fail[p];
113 }
114 }
115 puts("");
116 return 0;
117 }*/
118 for (R int i = 0; i < lb; ++i)
119 {
120 if (B[i] != ‘*‘) set(aph[B[i] - ‘a‘], i);
121 else for (R int j = 0; j < 26; ++j) set(aph[j], i);
122 set(ans, i);
123 }
124 for (R int i = 0; i < 26; ++i) init(i);
125 // fprintf(stderr, "%d %d\n", ‘*‘, ‘a‘);
126 for (R int i = 0; i < la; ++i) r[i] = i;
127 std::sort(r, r + la, cmp);
128 for (R int i = 0; i < la; ++i)
129 if (A[r[i]] != ‘*‘)
130 bit_and(A[r[i]] - ‘a‘, r[i]);
131 // ans &= (aph[A[i] - ‘a‘] >> i);
132 for (R int i = 0; i + la - 1 < lb; ++i) if (query(ans, i)) printf("%d ", i + 1); puts("");
133 return 0;
134 }
bitset
时间： 2024-11-05 14:38:14