hdu 1560 DNA sequence(搜索)

http://acm.hdu.edu.cn/showproblem.php?pid=1560

DNA sequence

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 732    Accepted Submission(s): 356

Problem Description

The twenty-first century is a biology-technology developing century. We know that a gene is made of DNA. The nucleotide bases from which DNA is built are A(adenine), C(cytosine), G(guanine), and T(thymine). Finding the longest common subsequence between DNA/Protein sequences is one of the basic problems in modern computational molecular biology. But this problem is a little different. Given several DNA sequences, you are asked to make a shortest sequence from them so that each of the given sequence is the subsequence of it.
For example, given "ACGT","ATGC","CGTT" and "CAGT", you can make a sequence in the following way. It is the shortest but may be not the only one.

Input

The first line is the test case number t. Then t test cases follow. In each case, the first line is an integer n ( 1<=n<=8 ) represents number of the DNA sequences. The following k lines contain the k sequences, one per line. Assuming that the length of any sequence is between 1 and 5.

Output

For each test case, print a line containing the length of the shortest sequence that can be made from these sequences.

Sample Input

1

4

ACGT
ATGC

CGTT

CAGT

Sample Output

8

Author

LL

参照大神代码优化hash标记节省时间1S以上，碉堡了：

有图有真相：

代码：

 #include<iostream>
 #include<stdio.h>
 #include<string.h>
 #include<queue>

 #define M 1679616+100000
 using namespace std;

 struct Nod
 {
     int st;
     int ln;
 }nd1,nd2;

 int hash[M];
 int n,len[],size,maks;
 char str[][];
 char ku[] ="ACGT";
 int cas =;  // hash标记用，这里可以节省大量初始化时间（大神专用，哈哈，看的大神代码都这么玩的）

 int bfs()
 {
     int pos[];
     queue<Nod> q;
     nd1.st = ;
     nd1.ln = ;
     q.push(nd1);
    // memset(hash,0,sizeof(hash));   //用上面的cas代替初始化过程，节省时间1s以上
     while(!q.empty())
     {
         nd2 = q.front();
         q.pop();
         int i,j;
         int st = nd2.st;
         for(i=;i<n;i++)
         {
             pos[i] = st%maks;
             st/=maks;
         }
         for(j=;j<;j++)
         {
             st = ;
             for(i=n-;i>=;i--)
             {
                 st = st * maks + pos[i] + (str[i][pos[i]] == ku[j]);
             }
             if(hash[st] == cas)    continue;
             if(st == size)    return nd2.ln + ;
             hash[st] = cas;
             nd1.ln = nd2.ln + ;
             nd1.st = st;
             q.push(nd1);
         }
     }
     return -;
 }

 int main()
 {
     int t;
     scanf("%d",&t);
     cas = ;
     while(t--)
     {
         scanf("%d",&n);
         int i;
         maks=;
         for(i=;i<n;i++)
         {
             scanf("%s",str[i]);
             len[i] = strlen(str[i]);
             if(maks<len[i]) maks = len[i];
         }
         maks++;
         size = ;
         for(i=n-;i>=;i--) size = size*maks+len[i];
         printf("%d\n",bfs());
         cas++;  //cas加1，用来标记下一组测试数据
     }
     return ;
 }