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

Input14ACGTATGCCGTTCAGT

Output8

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

123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657#include<bits/stdc++.h>#define INIT(a,b) memset(a,b,sizeof(a))#define LL long longusing 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;}