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[8],size,maks;
char str[8][8];
char ku[] ="ACGT";
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）

int bfs()
{
    int pos[8];
    queue<Nod> q;
    nd1.st = 0;
    nd1.ln = 0;
    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=0;i<n;i++)
        {
            pos[i] = st%maks;
            st/=maks;
        }
        for(j=0;j<4;j++)
        {
            st = 0;
            for(i=n-1;i>=0;i--)
            {
                st = st * maks + pos[i] + (str[i][pos[i]] == ku[j]);
            }
            if(hash[st] == cas)    continue;
            if(st == size)    return nd2.ln + 1;
            hash[st] = cas;
            nd1.ln = nd2.ln + 1;
            nd1.st = st;
            q.push(nd1);
        }
    }
    return -1;
}

int main()
{
    int t;
    scanf("%d",&t);
    cas = 1;
    while(t--)
    {
        scanf("%d",&n);
        int i;
        maks=0;
        for(i=0;i<n;i++)
        {
            scanf("%s",str[i]);
            len[i] = strlen(str[i]);
            if(maks<len[i]) maks = len[i];
        }
        maks++;
        size = 0;
        for(i=n-1;i>=0;i--) size = size*maks+len[i];
        printf("%d
",bfs());
        cas++;  //cas加1，用来标记下一组测试数据
    }
    return 0;
}