POJ 2778 DNA Sequence

DNA Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10868		Accepted: 4146

DescriptionIt's well known that DNA Sequence is a sequence only contains A, C, T and G, and it's very useful to analyze a segment of DNA Sequence，For example, if a animal's DNA sequence contains segment ATC then it may mean that the animal may have a genetic disease. Until now scientists have found several those segments, the problem is how many kinds of DNA sequences of a species don't contain those segments.Suppose that DNA sequences of a species is a sequence that consist of A, C, T and G，and the length of sequences is a given integer n. InputFirst line contains two integer m (0 <= m <= 10), n (1 <= n <=2000000000). Here, m is the number of genetic disease segment, and n is the length of sequences.Next m lines each line contain a DNA genetic disease segment, and length of these segments is not larger than 10. OutputAn integer, the number of DNA sequences, mod 100000.Sample Input

4 3 AT AC AG AA

Sample Output

36

题意：给出一组模式串求出给定长度的字符串的组合数，要满足不包含模式串。

sl:先将模式串建成一个自动机，（要把失配边加上）然后根据图构造出一个矩阵，求长度为k的字符串就相当于在自动机上走K步

n有2*1e9那么大不能用白书介绍的dp方法，想到离散数学，很容易知道，求长度为k的路径就是求矩阵的k次方，所以做下快速幂

然后把结果加起来就可以。

  1 #include<cstdio>2 #include<cstring>3 #include<algorithm>4 #include<queue>5 using namespace std;6 typedef long long LL;7 const int MAX = 200;8 const int MOD = 100000;9 char p[MAX];10 LL map[MAX][MAX];11 LL ans[MAX][MAX],sz;12 LL z[MAX][MAX];13 struct AC14 {15     int ch[MAX][MAX],val[MAX],f[MAX];16     void init(){ sz=1; memset(ch,0,sizeof(ch)); memset(val,0,sizeof(val));}1718     int index(char c)19     {20         if(c=='A') return 0;21         else if(c=='C') return 1;22         else if(c=='G') return 2;23         else return 3;24     }2526     void insert(char *s)27     {28         int u = 0, n = strlen(s);29         for(int i = 0; i < n; i++)30         {31             int c = index(s[i]);32             if(!ch[u][c])33             {34                 memset(ch[sz], 0, sizeof(ch[sz]));35                 val[sz] = 0;36                 ch[u][c] = sz++;37             }38             u = ch[u][c];39         }40         val[u] = 1;41       }4243     void getfail()44     {45         queue<int> q;46         f[0] = 0;47         for(int c = 0; c < 4; c++)48         {49           int u = ch[0][c];50           if(u) { f[u] = 0; q.push(u); }51         }52         while(!q.empty())53         {54           int r = q.front(); q.pop();55           for(int c = 0; c < 4; c++) {56             int u = ch[r][c];57             if(!u) { ch[r][c] = ch[f[r]][c]; continue; }58             q.push(u);59             int v = f[r];60             while(v && !ch[v][c]) v = f[v];61             f[u] = ch[v][c];62             val[u] |= val[f[u]];63           //  printf("%d\n",val[u]);64           }65         }66   }67 };686970 void multiply(LL x[MAX][MAX],LL y[MAX][MAX])71 {72     for(int i=0;i<sz;i++)73     {74         for(int j=0;j<sz;j++)75         {76             z[i][j]=0;77             for(int k=0;k<sz;k++)78             z[i][j]=z[i][j]+x[i][k]*y[k][j];79             z[i][j]%=MOD;80         }81     }82     for(int i=0;i<sz;i++)83     for(int j=0;j<sz;j++)84     y[i][j]=z[i][j];85 }86 int main()87 {88     int n,m;89     AC ac;90     while(scanf("%d %d",&m,&n)==2)91     {92         ac.init();93         for(int i=0;i<m;i++)94         {95             scanf("%s",p);96             ac.insert(p);97         }98         ac.getfail();99         memset(map,0,sizeof(map));memset(ans,0,sizeof(ans));for(int i=0;i<sz;i++){if(!ac.val[i]){for(int j=0;j<4;j++){if(!ac.val[ac.ch[i][j]])map[i][ac.ch[i][j]]++;}}}for(int i=0;i<sz;i++) ans[i][i]=1;while(n){if(n & 1) multiply(map,ans);multiply(map,map);n>>=1;}LL res = 0;for(int i=0;i<sz;i++) res=res+ans[0][i];printf("%I64d\n",res%MOD);return 0;}}