Poj 2778 DNA Sequence

题目链接：http://poj.org/problem?id=2778

题目类型：AC自动机 + 矩阵快速幂+ DP

题意:给定一些病毒的模式串，问给定的长度为N的所有DNA串中有多少是不含有病毒串的。

数据范围N最大为2000000000

这道题的解题思路可以参考：http://blog.henix.info/blog/poj-2778-aho-corasick-dp.html

和：http://www.matrix67.com/blog/archives/276

关键是AC自动机中失效函数要稍作变化。（我理解的失效函数有好几种写法，先记录三种，在注释中，依据问题不同而选用不同的方式)

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <math.h>#include <map>#include <queue>#include <algorithm>using namespace std;#define SIGMA_SIZE 4#define MAXNODE 105#define TEXT_SIZE 2000#define P_SIZE 15#define P_NUM 15#define MATRIX_SIZE 105#define MOD 100000struct Matrix{   long long int elem[MATRIX_SIZE][MATRIX_SIZE];   int size;   Matrix(){memset(elem,0,sizeof(elem));}   void setSize(int _size)   {      size = _size;   }   Matrix operator = (const Matrix & other)   {      setSize(other.size);      for(int i=0;i<size;i++)      {         for(int j= 0;j<size;j++)         {            elem[i][j] = other.elem[i][j];         }      }      return *this;   }   Matrix operator * (const Matrix & other)   {      Matrix temp;      temp.setSize(size);      for(int i=0;i<size;i++)      {         for(int j=0;j<size;j++)         {            for(int k=0;k<size;k++)            {               temp.elem[i][j] += elem[i][k] * other.elem[k][j];               if(temp.elem[i][j]>=MOD) temp.elem[i][j] %= MOD;            }         }      }      return temp;   }   void Power(int exp)   {      Matrix E;      E.setSize(size);      for(int i=0;i<size;i++) E.elem[i][i] = 1;      while(exp)      {         if(exp & 1) E = E * (*this);         *this = (*this) * (*this);         exp >>= 1;      }      *this = E;   }};struct AhoCorsickAutomata{   int cnt[P_NUM];   int sz;   int ch[MAXNODE][SIGMA_SIZE];   int f[MAXNODE];   int val[MAXNODE];   int last[MAXNODE];   void init()   {      sz = 1;      memset(ch[0],0,sizeof(ch[0]));      memset(cnt,0,sizeof(cnt));      memset(f,0,sizeof(f));   }   int idx(char c)   {      if(c == 'A') return 0;      if(c == 'C') return 1;      if(c == 'G') return 2;      if(c == 'T') return 3;   }   void insert(char *s,int v)   {      int u = 0,n = strlen(s);      for(int i=0; i<n; i++)      {         int c = idx(s[i]);         if(!ch[u][c])         {            memset(ch[sz],0,sizeof(ch[sz]));            val[sz] = 0;            ch[u][c] = sz++;         }         u = ch[u][c];         }         val[u] = v;      }   void print(int j)   {      if(j)      {         cnt[val[j]]++;         print(last[j]);      }   }   void find(char *T)   {      int n = strlen(T);      int j = 0;      for(int i=0; i<n; i++)      {         int c = idx(T[i]);         while(j && !ch[j][c]) j = f[j];         j = ch[j][c];         if(val[j]) print(j);         else if(last[j]) print(last[j]);      }   }   void getFail()   {      queue<int> q;      f[0] = 0;      for(int c = 0; c<SIGMA_SIZE; c++)      {         int u = ch[0][c];         if(u)         {            f[u] = 0;            q.push(u);            last[u] = 0;         }      }      while(!q.empty())      {         int r = q.front();         q.pop();         for(int c = 0; c<SIGMA_SIZE; c++)         {            /*The first way:the last() funciton exists and we will find repeatedly*/            /*int u = ch[r][c];            if(!u) continue;            q.push(u);            int v = f[r];            while(v && !ch[v][c]) v = f[v];            f[u] = ch[v][c];            last[u] = val[f[u]] ? f[u] : last[f[u]];            */            /*The second way:the last() function exists                but we see the all transfer in the same method            */            /*int u = ch[r][c];            if(!u)            {               ch[r][c] = ch[f[r]][c];               continue;            }            q.push(u);            int v = f[r];            f[u] = ch[v][c];            last[u] = val[f[u]] ? f[u] : last[f[u]];               */            /*The third way: we ingore the last() function                because we only foucus whether the current node               is the key node            */            ///*            int u = ch[r][c];            if(!u)            {               ch[r][c] = ch[f[r]][c];               continue;            }            q.push(u);            int v = f[r];            f[u] = ch[v][c];            val[u] |= val[f[u]];            //*/         }      }   }   Matrix work()   {      Matrix temp;      temp.setSize(sz);      for(int u=0;u<sz;u++)      {         if(val[u]) continue;         for(int j=0;j<SIGMA_SIZE;j++)         {            int v = ch[u][j];            if(val[v]) continue;            temp.elem[u][v] += 1;         }      }      return temp;   }};AhoCorsickAutomata ac;char text[TEXT_SIZE];char p[P_SIZE];int main(){   #ifndef ONLINE_JUDGE      freopen("in.txt","r",stdin);   #endif   int m,n;   while(scanf(" %d %d",&m,&n)!=EOF)   {      ac.init();      for(int i=1;i<=m;i++)      {         scanf(" %s",p);         ac.insert(p,1);      }      ac.getFail();      Matrix mat = ac.work();      mat.Power(n);      long long int ans = 0;      for(int i=0;i<mat.size;i++) ans += mat.elem[0][i];      printf("%lld\n",ans % MOD );   }   return 0;}