hdu 1560 DNA sequence(迭代加深搜索)(经典题)

DNA sequence

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1296    Accepted Submission(s): 648

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

14ACGTATGCCGTTCAGT

 

Sample Output

8

题意：

从ｎ个串中找出一个最短的公共串,，该公共串对于n个字符串不要求连续，即只要保持相对顺序就好。

思路：

用迭代加深搜索，所谓迭代加深搜索，就是限制DFS的深度，若搜不到答案，则加深深度，重新搜索，这样就防止了随着深度不断加深而进行的盲目搜索，而且，对于这种求最短长度之类的题目，只要找到可行解，即是最优解了。同时注意剪枝，每次DFS的时候，都要判断一下，当前的深度＋最少还有加深的深度是否大于限制的长度，若是，则退回上一层状态。

代码：

//1606MS1684K#include<iostream>#include<cstring>#include<algorithm>using namespace std;char str[10][10];//记录n个字符串int n,ans,deep,size[10];char DNA[4]={'A','C','G','T'};//一共四种可能void dfs(int cnt,int len[]){    if(cnt > deep)//大于限制的深度，不用往下搜索        return;    int maxx = 0;//预计还要匹配的字符串的最大长度    for(int i=0;i<n;i++)    {        int t = size[i]-len[i];        if(t>maxx)            maxx = t;    }    if(maxx==0)//条件全部满足即为最优解    {        ans = cnt;        return;    }    if(cnt + maxx > deep)        return;    for(int i=0;i<4;i++)    {        int pos[10];        int flag = 0;        for(int j=0;j<n;j++)        {            if(str[j][len[j]]==DNA[i])            {                flag = 1;                pos[j] = len[j]+1;            }            else                pos[j]=len[j];        }        if(flag)            dfs(cnt+1,pos);        if(ans!=-1)            break;    }}int main(){    int t;    cin>>t;    while(t--)    {        cin>>n;        int maxn = 0;        for(int i=0;i<n;i++)        {            cin>>str[i];            size[i] = strlen(str[i]);            if(size[i]>maxn)                maxn = size[i];        }        ans = -1;        deep = maxn;        int pos[10] = {0};//记录n个字符串目前匹配到的位置        while(1)        {            dfs(0,pos);            if(ans!=-1)                break;            deep++;//加深迭代        }        cout<<ans<<endl;    }    return 0;}