标签:gcd size 快速幂 根据 val 表示 linker space pair
题意:给定一些串,然后让你构造出一个长度为 m 的串,并且不包含以上串,问你有多少个。
析:很明显,如果 m 小的话 ,直接可以用DP来解决,但是 m 太大了,我们可以认为是在AC自动机图中,根据离散中的矩阵的幂可以表示 从 i 到 j 需要 x 步的有多少条。比如A[1][2]^5 = 10,表示从结点 1 到结点 2 走五步有10种方法,利用这种方法,就可以直接进行矩阵快速幂了。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #include <list> #include <assert.h> #include <bitset> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define fi first #define se second #define pb push_back #define sqr(x) ((x)*(x)) #define ms(a,b) memset(a, b, sizeof a) //#define sz size() #define pu push_up #define pd push_down #define cl clear() #define all 1,n,1 #define FOR(i,x,n) for(int i = (x); i < (n); ++i) #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 1e20; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1000 + 10; const int maxm = 1e5 + 10; const int mod = 100000; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, -1, 0, 1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c) { return r >= 0 && r < n && c >= 0 && c < m; } const int maxnode = 10 * 10 + 10; const int sigma = 4; struct Matrix{ int n; int a[maxnode][maxnode]; void clear(){ ms(a, 0); } friend Matrix operator * (const Matrix &lhs, const Matrix &rhs){ Matrix res; res.n = lhs.n; FOR(i, 0, lhs.n) for(int j = 0; j < lhs.n; ++j){ res.a[i][j] = 0; for(int k = 0; k < lhs.n; ++k) res.a[i][j] = (res.a[i][j] + (LL)lhs.a[i][k] * rhs.a[k][j]) % mod; } return res; } }; Matrix x; struct Aho{ int ch[maxnode][sigma]; int f[maxnode]; bool val[maxnode]; int sz; void clear(){ sz = 1; ms(ch[0], 0); } inline int idx(char ch){ if(ch == ‘A‘) return 0; if(ch == ‘C‘) return 1; if(ch == ‘G‘) return 2; return 3; } void insert(const char *s){ int u = 0; while(*s){ int c = idx(*s); if(!ch[u][c]){ ms(ch[sz], 0); val[sz] = 0; ch[u][c] = sz++; } u = ch[u][c]; ++s; } val[u] = 1; } void getFail(){ queue<int> q; f[0] = 0; for(int c = 0; c < sigma; ++c){ int u = ch[0][c]; if(u){ f[u] = 0; q.push(u); } } while(!q.empty()){ int r = q.front(); q.pop(); for(int c = 0; c < sigma; ++c){ int u = ch[r][c]; if(!u){ ch[r][c] = ch[f[r]][c]; continue; } q.push(u); int v = f[r]; while(v && !ch[v][c]) v = f[v]; f[u] = ch[v][c]; val[u] |= val[f[u]]; } } } void solve(){ for(int i = 0; i < sz; ++i) if(!val[i]){ for(int j = 0; j < sigma; ++j){ int u = ch[i][j]; if(!val[u]) ++x.a[i][u]; } } x.n = sz; } }; Aho aho; char s[maxnode]; Matrix fast_pow(Matrix x, int n){ Matrix res; res.cl; res.n = x.n; for(int i = 0; i < res.n; ++i) res.a[i][i] = 1; while(n){ if(n&1) res = res * x; n >>= 1; x = x * x; } return res; } int main(){ while(scanf("%d %d", &n, &m) == 2){ aho.cl; for(int i = 0; i < n; ++i){ scanf("%s", s); aho.insert(s); } aho.getFail(); x.cl; aho.solve(); Matrix res = fast_pow(x, m); int ans = 0; for(int i = 0; i < res.n; ++i) ans = (ans + res.a[0][i]) % mod; printf("%d\n", ans); } return 0; }
POJ 2778 DNA Sequence (AC自动机+DP+矩阵)
标签:gcd size 快速幂 根据 val 表示 linker space pair
原文地址:http://www.cnblogs.com/dwtfukgv/p/7633271.html