Zoj 3494 BCD Code (AC automation + digital dp)
Link: zoj 3494 BCD Code
Given n binary strings, then there is a range of l, r. Q: l, the number of r pairs converted to BCD binary does not contain the above binary string.
Solution: AC automation + digital dp. First, establish an AC automatic mechanism for the disallow string. All word nodes are nodes that are not allowed to pass. Then perform the digital dp,
Use solve (r)-solve L-1 1). Here, l requires the subtraction of large numbers. Dp [I] [j] indicates the number of feasible solutions for moving I to node j. Each time the next digit is enumerated
It is BCD binary, so each digit needs to be moved 4 characters at a time. If there is a prohibited point in the middle, it will not work. Then, use an eq and mv to mark equal conditions.
If the conditions are equal, it may not be possible, that is, when the eq is 0. Note the handling of leading 0, because some prohibited strings can be composed of 0, so for the first
0 cannot be moved on the AC automatic machine, so it is taken into consideration separately when each digit is used.
#include
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#include using namespace std;const int maxn = 2005;const int mod = 1e9 + 9;const int sigma_size = 2;const char sign[12][5] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001"};struct Aho_Corasick { int sz, g[maxn][sigma_size]; int tag[maxn], fail[maxn], last[maxn]; int dp[205][maxn]; void init(); int idx(char ch); void insert(char* str, int k); void getFail(); void match(char* str); void put(int x, int y); int solve(int* a, int n); int move(int u, int v);}AC;int N, num[205];int getNum(int* a) { char s[205]; scanf("%s", s); int n = strlen(s), mv = 0; while (mv < n && s[mv] == '0') mv++; n -= mv; if (n == 0) num[n++] = 0; else { for (int i = 0; i < n; i++) num[i] = s[i+mv] - '0'; } return n;}void del(int* a, int& n) { if (n == 0) return; a[n-1]--; for (int i = n-1; i >= 0; i--) { if (a[i] < 0) { a[i] += 10; a[i-1]--; } else break; } if (a[0] == 0) { for (int i = 0; i < n; i++) a[i] = a[i+1]; n--; }}int main () { int cas, n; char w[50]; scanf("%d", &cas); while (cas--) { AC.init(); scanf("%d", &n); for (int i = 1; i <= n; i++) { scanf("%s", w); AC.insert(w, i); } AC.getFail(); N = getNum(num); del(num, N); int l = AC.solve(num, N); N = getNum(num); int r = AC.solve(num, N); printf("%d\n", (r - l + mod) % mod); } return 0;}int Aho_Corasick::move(int u, int id) { for (int i = 0; i < 4; i++) { int v = idx(sign[id][i]); while (u && g[u][v] == 0) u = fail[u]; u = g[u][v]; if (tag[u] || last[u]) return -1; } return u;}int Aho_Corasick::solve(int* a, int n) { if (n == 0) return 0; memset(dp, 0, sizeof(dp)); int eq = 1, mv = 0; for (int i = 0; i < n; i++) { for (int x = 0; x < sz; x++) { if (tag[x] || last[x]) continue; for (int j = 0; j < 10; j++) { int u = move(x, j); if (u == -1) continue; dp[i+1][u] = (dp[i+1][u] + dp[i][x]) % mod; } } if (i) { for (int j = 1; j < 10; j++) { int u = move(0, j); if (u == -1) continue; dp[i+1][u] = (dp[i+1][u] + 1) % mod; } } if (eq) { for (int j = (i == 0 ? 1 : 0); j < a[i]; j++) { int u = move(mv, j); if (u == -1) continue; dp[i+1][u] = (dp[i+1][u] + 1) % mod; } int u = move(mv, a[i]); if (u == -1) eq = 0; mv = u; } } int ans = eq; for (int i = 0; i < sz; i++) { if (tag[i] || last[i]) continue; ans = (ans + dp[n][i]) % mod; } return ans;}void Aho_Corasick::init() { sz = 1; tag[0] = 0; memset(g[0], 0, sizeof(g[0]));}int Aho_Corasick::idx(char ch) { return ch - '0';}void Aho_Corasick::put(int x, int y) {}void Aho_Corasick::insert(char* str, int k) { int u = 0, n = strlen(str); for (int i = 0; i < n; i++) { int v = idx(str[i]); if (g[u][v] == 0) { tag[sz] = 0; memset(g[sz], 0, sizeof(g[sz])); g[u][v] = sz++; } u = g[u][v]; } tag[u] = k;}void Aho_Corasick::match(char* str) { int n = strlen(str), u = 0; for (int i = 0; i < n; i++) { int v = idx(str[i]); while (u && g[u][v] == 0) u = fail[u]; u = g[u][v]; if (tag[u]) put(i, u); else if (last[u]) put(i, last[u]); }}void Aho_Corasick::getFail() { queue
que; for (int i = 0; i < sigma_size; i++) { int u = g[0][i]; if (u) { fail[u] = last[u] = 0; que.push(u); } } while (!que.empty()) { int r = que.front(); que.pop(); for (int i = 0; i < sigma_size; i++) { int u = g[r][i]; if (u == 0) { g[r][i] = g[fail[r]][i]; continue; } que.push(u); int v = fail[r]; while (v && g[v][i] == 0) v = fail[v]; fail[u] = g[v][i]; last[u] = tag[fail[u]] ? fail[u] : last[fail[u]]; } }}