HDU-1560 DNA sequence（IDA*）

DNA sequence

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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
   
   
   
   

数据不是很大，AC后又进入前25了。

开始TLE 5次，一直以为是自己的估价函数没写好，不停的改成现在这个样子，结果还是TLE，百度到大神的代码后发现他的优化比我还少，就想到了一定是初始化的问题，改过之后立刻AC，debug真是一个艰辛的过程。

经过测试后，发现如果只有以DNA最大剩余长度为估价函数返回值，几乎要TLE，所以主要起作用的是剩余DNA中ACGT分别出现的次数的最大值的和

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

char s[41],dna[9][6],nuc[]={"ACGT"};
int n,len[9],num[9],pre[41][9],step,depth,tp;
int vis[255],mx[255],mxl;
inline bool judge() {//判断是否全部完成匹配
    for(int i=0;i<n;++i)
        if(len[i]!=num[i])
            return false;
    return true;
}

int get_h() {//估价函数：返回（未匹配DNA的最大长度）与（剩余DNA中ACGT分别出现的次数的最大值的和）的最大值
    int i,j;
    mx['A']=mx['C']=mx['G']=mx['T']=mxl=0;
    for(i=0;i<n;++i) {
        vis['A']=vis['C']=vis['G']=vis['T']=0;
        for(j=num[i];j<len[i];++j)
            ++vis[dna[i][j]];//每一个剩余DNA中ACGT分别出现的次数
        mx['A']=max(mx['A'],vis['A']);
        mx['C']=max(mx['C'],vis['C']);
        mx['G']=max(mx['G'],vis['G']);//剩余DNA中ACGT分别出现的次数的最大值
        mx['T']=max(mx['T'],vis['T']);
        mxl=max(mxl,len[i]-num[i]);//DNA剩余的长度的最大值
    }
    return max(mx['A']+mx['C']+mx['G']+mx['T'],mxl);
}

bool IDAstar() {
    if(judge())
        return true;
    if(step+get_h()>=depth)//剪枝
        return false;
    int i,j;
    for(i=0;i<4;++i) {
        s[step]=nuc[i];
        for(j=0;j<n;++j) {
            if(s[step]==dna[j][num[j]])
                num[j]+=(pre[step][j]=1);//DNA成功匹配一个字符
            else
                pre[step][j]=0;
            tp+=pre[step][j];//DNA成功匹配字符的个数
        }
        if(tp!=0) {//DNA至少有一个字符成功匹配时才进入下一层
            ++step;
            if(IDAstar())
                return true;
            --step;
        }
        for(j=0;j<n;++j)//DNA成功匹配的字符数复原之字符串到位置step的状态
            num[j]-=pre[step][j];
    }
    return false;
}

int main() {
    int i,T;
    scanf("%d",&T);
    while(T--) {
        scanf("%d",&n);
        depth=1;
        for(i=0;i<n;++i) {
            scanf("%s",dna[i]);
            depth=max(depth,len[i]=strlen(dna[i]));
            num[i]=0;
        }
        step=0;
        while(!IDAstar()) {
            ++depth;
        }
        printf("%d\n",step);
    }
    return 0;
}