DNA sequence HDU

DNA sequence

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4217    Accepted Submission(s): 2020

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

 

Source

HDU 2006-12 Programming Contest

 

Recommend

LL

 

题意：现在我们给定了数个DNA序列，请你构造出一个最短的DNA序列，使得所有我们给定的DNA序列都是它的子序列。例如，给定"ACGT","ATGC","CGTT","CAGT",你可以构造的一个最短序列为"ACAGTGCT"，但是需要注意的是，这并不是此问题的唯一解。

思路： *IDA  若当前长度+最长构造长度< 迭代的最大深度则return 进行良好剪枝

accode

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<string>
#include<vector>
#include<set>
#include<stack>
#include<queue>
#include<map>
#include<cmath>
#include<algorithm>
using namespace std;
#define inf 0x3f3f3f3f
#define ll long long
#define MAX_N 1000005
#define gcd(a,b) __gcd(a,b)
#define mem(a,x) memset(a,x,sizeof(a))
#define mid(a,b) a+b/2
#define stol(a) atoi(a.c_str())//string to long
string temp[10];
string c = "ACGT";
int pos[10];
int len[10];
int deep;
int n;
int ans(){
    int maxn = 0;
    for(int i = 0; i < n; i++){
        maxn = max(maxn,len[i]-pos[i]);
    }
    return maxn;
}
int dfs(int step){
    if(step + ans() > deep)
        return 0;
    if(!ans())//满足构造序列包含所有序列
        return 1;
    int temp1[10];
    for(int i = 0; i < 4; i++){

        int flag = 0;
        for(int j = 0; j < n; j++)
            temp1[j] = pos[j];

        for(int j = 0; j < n; j++){
            if(temp[j][pos[j]] == c[i]){
                flag = 1;
                pos[j]++;
            }
        }
        if(flag){
            if(dfs(step+1))
                return 1;
            for(int j = 0; j < n; j++)
                pos[j] = temp1[j];
        }
    }
    return 0;
}
int main(){
//    freopen("D:\in.txt","r",stdin);
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        deep = 0;
        for(int i = 0; i < n; i++){
            cin>>temp[i];
            len[i] = temp[i].length();
            pos[i] = 0;
            deep = max(deep,len[i]);
        }
        while(1){
            if(dfs(0))
                break;
            ++deep;
        }
        printf("%d
",deep);
    }
    return 0;
}