【AC自动机+矩阵快速幂】 POJ 2778 DNA Sequence

先构建AC自动机,然后通过AC自动机构建矩阵,最后矩阵快速幂即可。。。

#include <iostream>  
#include <queue>  
#include <stack>  
#include <map>  
#include <set>  
#include <bitset>  
#include <cstdio>  
#include <algorithm>  
#include <cstring>  
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 300005
#define maxm 300005
#define eps 1e-10
#define mod 100000
#define INF 1e9
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid  
#define rson o<<1 | 1, mid+1, R  
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}
LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}
// head

struct AC
{
	int next[maxn][4];
	int fail[maxn];
	int end[maxn];
	char s[maxn];
	queue<int> q;
	int top, now, root;
	
	int newnode(void)
	{
		end[top] = 0;
		fail[top] = -1;
		for(int i = 0; i < 4; i++)
			next[top][i] = -1;
		return top++;
	}
	
	void init(void)
	{
		top = 0;
		root = newnode();
	}
	
	int hash(char ss)
	{
		if(ss == 'A') return 0;
		if(ss == 'C') return 1;
		if(ss == 'G') return 2;
		if(ss == 'T') return 3;
	}
	
	void insert(void)
	{
		int len = strlen(s);
		now = root;
		for(int i = 0; i < len; i++) {
			int t = hash(s[i]);
			if(next[now][t] == -1) next[now][t] = newnode();
			now = next[now][t];
		}
		end[now] = 1;
	}
	
	void build(void)
	{
		now = root;
		for(int i = 0; i < 4; i++)
			if(next[now][i] == -1) next[now][i] = root;
			else {
				fail[next[now][i]] = root;
				q.push(next[now][i]);
			}
		while(!q.empty()) {
			now = q.front(), q.pop();
			end[now] |= end[fail[now]];
			for(int i = 0; i < 4; i++)
				if(next[now][i] == -1) next[now][i] = next[fail[now]][i];
				else {
					fail[next[now][i]] = next[fail[now]][i];
					q.push(next[now][i]);
				}
		}
	}
};

struct matrix
{
	LL mat[105][105];
	LL res[105][105];
	LL mid[105][105];
	int n;
	
	void init(void)
	{
		memset(res, 0, sizeof res);
		for(int i = 0; i < n; i++) res[i][i] = 1;
		memset(mat, 0, sizeof mat);
	}
	
	void work(int c)
	{
		while(c) {
			if(c%2) {
				for(int i = 0; i < n; i++)
					for(int j = 0; j < n; j++) {
						LL tmp = 0;
						for(int k = 0; k < n; k++)
							tmp = (tmp + res[i][k] * mat[k][j]) % mod;
						mid[i][j] = tmp;
					}
				for(int i = 0; i < n; i++)
					for(int j = 0; j < n; j++)
						res[i][j] = mid[i][j];
			}
			for(int i = 0; i < n; i++)
				for(int j = 0; j < n; j++) {
					LL tmp = 0;
					for(int k = 0; k < n; k++)
						tmp = (tmp + mat[i][k] * mat[k][j]) % mod;
					mid[i][j] = tmp;
				}
			for(int i = 0; i < n; i++)
				for(int j = 0; j < n; j++)
					mat[i][j] = mid[i][j];
			c /= 2;
		}
	}
};

int n, m;
AC ac;
void read(void)
{
	ac.init();
	while(m--) {
		scanf("%s", ac.s);
		ac.insert();
	}
	ac.build();
}

matrix t;

void work(void)
{
	t.n = ac.top;
	t.init();
	for(int i = 0; i < ac.top; i++)
		for(int j = 0; j < 4; j++)
			if(!ac.end[ac.next[i][j]])
				t.mat[i][ac.next[i][j]]++;
	t.work(n);
	LL ans = 0;
	for(int i = 0; i < ac.top; i++) ans = (ans + t.res[0][i]) % mod;
	printf("%I64d\n", ans);
}

int main(void)
{
	while(scanf("%d%d", &m, &n)!=EOF) {
		read();
		work();
	}
	return 0;
}


你可能感兴趣的:(poj)