hdu 1560 DNA sequence(搜索)

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

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. hdu 1560 DNA sequence(搜索)

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

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：

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<string.h>
 4 #include<queue>
 5 
 6 #define M 1679616+100000
 7 using namespace std;
 8 
 9 struct Nod
10 {
11     int st;
12     int ln;
13 }nd1,nd2;
14 
15 int hash[M];
16 int n,len[8],size,maks;
17 char str[8][8];
18 char ku[] ="ACGT";
19 int cas =1;  // 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）
20 
21 int bfs()
22 {
23     int pos[8];
24     queue<Nod> q;
25     nd1.st = 0;
26     nd1.ln = 0;
27     q.push(nd1);
28    // memset(hash,0,sizeof(hash));   //用上面的cas代替初始化过程，节省时间1s以上
29     while(!q.empty())
30     {
31         nd2 = q.front();
32         q.pop();
33         int i,j;
34         int st = nd2.st;
35         for(i=0;i<n;i++)
36         {
37             pos[i] = st%maks;
38             st/=maks;
39         }
40         for(j=0;j<4;j++)
41         {
42             st = 0;
43             for(i=n-1;i>=0;i--)
44             {
45                 st = st * maks + pos[i] + (str[i][pos[i]] == ku[j]);
46             }
47             if(hash[st] == cas)    continue;
48             if(st == size)    return nd2.ln + 1;
49             hash[st] = cas;
50             nd1.ln = nd2.ln + 1;
51             nd1.st = st;
52             q.push(nd1);
53         }
54     }
55     return -1;
56 }
57 
58 int main()
59 {
60     int t;
61     scanf("%d",&t);
62     cas = 1;
63     while(t--)
64     {
65         scanf("%d",&n);
66         int i;
67         maks=0;
68         for(i=0;i<n;i++)
69         {
70             scanf("%s",str[i]);
71             len[i] = strlen(str[i]);
72             if(maks<len[i]) maks = len[i];
73         }
74         maks++;
75         size = 0;
76         for(i=n-1;i>=0;i--) size = size*maks+len[i];
77         printf("%d\n",bfs());
78         cas++;  //cas加1，用来标记下一组测试数据
79     }
80     return 0;
81 }