[POJ2778]DNA Sequence(AC自动机 + DP + 矩阵优化)
Posted 蒟蒻zht的博客
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AC自动机加DP就不说了
注意到 m <= 10,所以模式串很少。
而 n 很大就需要 log 的算法,很容易想到矩阵。
但是该怎么构建?
还是矩阵 A(i,j) = ∑A(i,k) * A(k,j),那么i到j的方案数就是j到k的方案数称k到j的方案数,那么直接矩阵快速幂即可
#include <queue> #include <cstdio> #include <cstring> #define N 100001 #define p 100000 #define LL long long int n, m, cnt, ans; int next[N][4], fail[N], f[101][N]; bool val[N]; char s[N]; std::queue <int> q; struct Matrix { int n, m; LL a[141][141]; Matrix() { n = m = 0; memset(a, 0, sizeof(a)); } }res; inline int idx(char x) { if(x == ‘A‘) return 0; if(x == ‘C‘) return 1; if(x == ‘T‘) return 2; if(x == ‘G‘) return 3; } inline void insert() { int i, x, now = 0, len = strlen(s + 1); for(i = 1; i <= len; i++) { x = idx(s[i]); if(!next[now][x]) next[now][x] = ++cnt; now = next[now][x]; } val[now] = 1; } inline void make_fail() { int i, now; for(i = 0; i < 4; i++) if(next[0][i]) q.push(next[0][i]); while(!q.empty()) { now = q.front(); q.pop(); for(i = 0; i < 4; i++) { if(!next[now][i]) { next[now][i] = next[fail[now]][i]; continue; } fail[next[now][i]] = next[fail[now]][i]; val[next[now][i]] |= val[next[fail[now]][i]]; q.push(next[now][i]); } } } inline Matrix operator * (Matrix x, Matrix y) { int i, j, k; Matrix ret; ret.n = x.n; ret.m = y.m; for(i = 0; i <= x.n; i++) for(j = 0; j <= y.m; j++) for(k = 0; k <= y.n; k++) ret.a[i][j] = (ret.a[i][j] + x.a[i][k] * y.a[k][j]) % p; return ret; } inline Matrix operator ^ (Matrix x, int y) { int i; Matrix ret; res.n = ret.m = cnt; for(i = 0; i <= cnt; i++) ret.a[i][i] = 1; for(; y; y >>= 1) { if(y & 1) ret = ret * x; x = x * x; } return ret; } int main() { int i, j, k; scanf("%d %d", &n, &m); for(i = 1; i <= n; i++) { scanf("%s", s + 1); insert(); } make_fail(); res.n = res.m = cnt; for(i = 0; i <= cnt; i++) { if(val[i]) continue; for(j = 0; j < 4; j++) if(!val[next[i][j]]) res.a[i][next[i][j]]++; } res = res ^ m; for(i = 0; i <= cnt; i++) ans = (ans + res.a[0][i]) % p; printf("%d\n", ans); return 0; }
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