DNA sequence（映射+BFS）

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.

SampleInput

1
4
ACGT
ATGC
CGTT
CAGT

SampleOutput

8

題意就是給你幾個DNA序列，要求找到一個序列，使得全部序列都是它的子序列（不必定連續）。
直接搜MLE、TLE、RE，因此不能直接搜索，通常處理這種序列問題，都是把序列映射到整數或其餘便於處理的東西上。
題目還說了每一個DNA的序列長度不會超過5，因此咱們能夠按位處理映射到一個整數上，並且題目只須要咱們輸出最短的序列長度，因此咱們也沒必要去映射字符，映射長度便夠了。
最多8個字符，每一個字符1-5長度，因此最大數爲6^8。好爲何是6^8，不明明是5^8麼，這個我暫時先不解釋，我加在了代碼註釋裏。
代碼：

 1 #include <iostream>
 2 #include <string>
 3 #include <cstdio>
 4 #include <cstdlib>
 5 #include <sstream>
 6 #include <iomanip>
 7 #include <map>
 8 #include <stack>
 9 #include <deque>
 10 #include <queue>
 11 #include <vector>
 12 #include <set>
 13 #include <list>
 14 #include <cstring>
 15 #include <cctype>
 16 #include <algorithm>
 17 #include <iterator>
 18 #include <cmath>
 19 #include <bitset>
 20 #include <ctime>
 21 #include <fstream>
 22 #include <limits.h>
 23 #include <numeric>
 24 
 25 using namespace std;  26 
 27 #define F first
 28 #define S second
 29 #define mian main
 30 #define ture true
 31 
 32 #define MAXN 1000000+5
 33 #define MOD 1000000007
 34 #define PI (acos(-1.0))
 35 #define EPS 1e-6
 36 #define MMT(s) memset(s, 0, sizeof s)
 37 typedef unsigned long long ull;  38 typedef long long ll;  39 typedef double db;  40 typedef long double ldb;  41 typedef stringstream sstm;  42 const int INF = 0x3f3f3f3f;  43 
 44 int t,n;  45 map<int,int>vis;  46 char s[10][10]; //保存序列  47 int len[10]; //保存每一個序列的長度  48 int p[10] = {1,6,36,216,1296,7776,46656,279936,1679616,10077696}; //6的k次方表  49 char temp[4]={'A','C','G','T'};  50 
 51 struct node{  52     int step; //長度  53     int st; //也就是映射數  54  node(){}  55     node(int _step, int _st):step(_step),st(_st){}  56 };  57 
 58 int bfs(int res){  59  vis.clear();  60     queue<node>q;  61     q.push(node(0,0));  62     vis[0] = 1;  63     while(!q.empty()){  64         node nxt,k = q.front();  65  q.pop();  66         if(k.st == res){ //當映射等於結果時 返回長度  67             return k.step;  68  }  69         for(int i = 0; i < 4; i++){  70             nxt.st = 0;  71             nxt.step = k.step+1;  72             int tp = k.st;  73             for(int j = 1; j <= n; j++){  74                 int x = tp%6; //獲得位數  75                 tp /= 6;  76                 if(x == len[j] || s[j][x+1] != temp[i]){ //判斷字符是否匹配  77                     nxt.st += x*p[j-1];  78  }  79                 else{  80                     nxt.st += (x+1)*p[j-1];  81  }  82  }  83             if(vis[nxt.st] == 0){ //標記是否已經搜過  84  q.push(nxt);  85                 vis[nxt.st] = 1;  86  }  87  }  88  }  89 }  90 
 91 int main(){  92     ios_base::sync_with_stdio(false);  93     cout.tie(0);  94     cin.tie(0);  95     cin>>t;  96     while(t--){  97         cin>>n;  98         int res = 0;  99         for(int i = 1; i <= n; i++){ //由於數組從0開始計數，但咱們映射以及後面操做都是基於位置，因此從1開始 100             cin>>s[i]+1; //同理從一開始 101             len[i] = strlen(s[i]+1); 102             res += len[i]*p[i-1]; //這也就是爲何是6^8，由於咱們是從1開始有5個狀態而不是0 103  } 104         cout << bfs(res) <<endl; 105  } 106     return 0; 107 }

因此這題你非要從0位置搞，弄5^8確實沒錯，也能夠作出來，可是操做會繁瑣不少，還不如從方便的角度多加一個長度。

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這道題的難度就是不知道怎麼入手，即便知道轉換處理也不知道該如何轉換以及如何搜索，這裏咱們避免了去從字符開始搜索，而是直接基於長度搜。

值得一提的是，我問了隊友後，他們表示這道題作法不少，還能夠用IDA*算法或者啓發式搜索，甚至不用搜索用AC自動機加矩陣也能夠作。但這些作法都是基於字符去搜索的，也不能說誰好誰壞，只是咱們的思惟就不同了，不少題目其實都不止一種解法，多想一想，頗有用的。至於其餘作法我也就懶得作了（實際上是不會23333）