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 题意就是给你几个DNA序列，要求找到一个序列，使得所有序列都是它的子序列（不一定连续）。直接搜MLE、TLE、RE，所以不能直接搜索，一般处理这种序列问题，都是把序列映射到整数或其他便于处理的东西上。题目还说了每个DNA的序列长度不会超过5，所以我们可以按位处理映射到一个整数上，而且题目只需要我们输出最短的序列长度，所以我们也不必去映射字符，映射长度便够了。最多8个字符，每个字符1-5长度，所以最大数为6^8。好为什么是6^8，不明明是5^8么，这个我暂时先不解释，我加在了代码注释里。代码：
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]; //保存每个序列的长度
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; //长度
53 int st; //也就是映射数
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){ //当映射等于结果时 返回长度
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; //得到位数
75 tp /= 6;
76 if(x == len[j] || s[j][x+1] != temp[i]){ //判断字符是否匹配
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){ //标记是否已经搜过
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++){ //因为数组从0开始计数，但我们映射以及后面操作都是基于位置，所以从1开始
100 cin>>s[i]+1; //同理从一开始
101 len[i] = strlen(s[i]+1);
102 res += len[i]*p[i-1]; //这也就是为什么是6^8，因为我们是从1开始有5个状态而不是0
103 }
104 cout << bfs(res) <<endl;
105 }
106 return 0;
107 }
所以这题你非要从0位置搞，弄5^8确实没错，也可以做出来，但是操作会繁琐很多，还不如从方便的角度多加一个长度。
这道题的难度就是不知道怎么入手，即使知道转换处理也不知道该如何转换以及如何搜索，这里我们避免了去从字符开始搜索，而是直接基于长度搜。
值得一提的是，我问了队友后，他们表示这道题做法很多，还可以用IDA*算法或者启发式搜索，甚至不用搜索用AC自动机加矩阵也可以做。但这些做法都是基于字符去搜索的，也不能说谁好谁坏，只是我们的思维就不一样了，很多题目其实都不止一种解法，多想想，很有用的。至于其他做法我也就懒得做了（其实是不会23333）
原文地址：https://www.cnblogs.com/xenny/p/9388400.html