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：
 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 }

 

posted @ 2013-07-26 21:44  crazy_apple  阅读(948)  评论(0)  编辑  收藏  举报