HDU1560(迭代加深搜索)

DNA sequence

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1914    Accepted Submission(s): 946

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

思路：迭代加深搜索。含义：在不知道迭代深度的前提下，依次探索每次的搜索深度。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN=15;
struct Node{
    char s[MAXN];
    int len,top;
}DNA[MAXN];
int n;
char buf[4]={'A','C','G','T'};
bool dfs(int l,int limit)
{
    int remain=0;
    for(int i=0;i<n;i++)
    {
        if(DNA[i].len-DNA[i].top>remain)
        {
            remain=DNA[i].len-DNA[i].top;
        }
    }
    if(remain==0)    return true;
    if(remain+l>limit)    return false; //重要剪枝一 
    
    for(int i=0;i<4;i++)
    {
        bool tag=false;
        int vis[MAXN]={0};
        for(int j=0;j<n;j++)    
        {
            if(buf[i]==DNA[j].s[DNA[j].top])
            {
                vis[j]=1;
                DNA[j].top++;
                tag=true;
            }
        }
        if(!tag)    continue;//重要剪枝二 
        if(dfs(l+1,limit))
        {
            return true;
        }
        for(int j=0;j<n;j++)
        {
            if(vis[j])
            {
                DNA[j].top--;
            }
        }
    }
    return false;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        int limit=0;
        for(int i=0;i<n;i++)
        {
            scanf("%s",DNA[i].s);
            DNA[i].len=strlen(DNA[i].s);        
            DNA[i].top=0;
            limit=max(limit,DNA[i].len);
        }
        while(!dfs(0,limit))
        {
            for(int i=0;i<n;i++)
            {
                DNA[i].top=0;
            }
            limit++;
        }
        printf("%d
",limit);
    }
    return 0;
}