DNA Sequence(IDA*,迭代深搜)

传送门HDU1560

描述

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.

输入

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.

输出

For each test case, print a line containing the length of the shortest sequence that can be made from these sequences.

样例

  • Input
    1
    4
    ACGT
    ATGC
    CGTT
    CAGT

  • Output
    8

题解

  • 题意:给出几个DNA序列,求一个最短的字符串,使得这些DNA序列都是它的子序列
  • 此题用裸dfs的话不知道结束条件容易超时,用裸bfs很容易超内存,所以我们采用迭代深搜的方法。所谓迭代深搜就是限制dfs的深度,逐渐将深度加深,起到模拟bfs的作用,又不需要bfs那么多内存,但是肯定是比bfs耗时要长
  • 在迭代深搜时,用deep表示限制深度,用tot表示当前串长度,用pos[i]表示i串目前已被匹配的字符个数,如果(tot+(len[i]-pos[i])>deep)的话,说明当前限制下的深度肯定不够,这样起到剪枝的效果,也就成了所谓的IDA*搜索

Code

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#include<bits/stdc++.h>
#define INIT(a,b) memset(a,b,sizeof(a))
#define LL long long
using namespace std;
const int inf=0x3f3f3f3f;
const int N=1e6+7;
const int mod=1e9+7;
string dna[10];
int len[10],pos[10];
string s="ACGT";
int n,deep;
int fut(){
int ans=0;
for(int i=0;i<n;i++)
ans=max(ans,len[i]-pos[i]);
return ans;
}
int dfs(int tot){
if(tot+fut()>deep) return 0;
if(!fut()) return 1;
int tem[10],flag=0;
for(int i=0;i<4;i++){
flag=0;
for(int j=0;j<n;i++)
tem[i]=pos[i];
for(int j=0;j<n;j++){
if(s[i]==dna[j][pos[j]]){
pos[j]++;
flag=1;
}
}
if(flag){
if(dfs(tot+1))return 1;
for(int i=0;i<n;i++)
pos[i]=tem[i];
}
}
return 0;
}
int main(){
int t;
cin>>t;
while(t--){
deep=0;
cin>>n;
for(int i=0;i<n;i++){
cin>>dna[i];
len[i]=dna[i].size();
pos[i]=0;
deep=max(deep,len[i]);
}
while(!dfs(0))
deep++;
cout<<deep<<endl;
}
return 0;
}
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