题意：两端基因片段，各有明确的碱基序列，现有一个碱基匹配的相似度数组，设计程序使得该相似度最大。

 1 //POJ1080-ZOJ1027
 2 //题解：将s1碱基和s2碱基看做等长，添加一个碱基为'-'，即每次都将上下碱基相互匹配后再和'-'匹配，找到最优解即可
 3 //即分为三种状态：
 4 //    1.s1[i]和s2[j]匹配
 5 //    2.s1[i]和 '-' 匹配
 6 //    3. '-' 和s2[j]匹配
 7 #include <iostream>
 8 #include <cstdio>
 9 #include <map>
10 using namespace std;
11 
12 #define MAX 105
13 #define max(x,y) ((x)>(y)?(x):(y))
14 
15 char s1[MAX], s2[MAX];
16 int len1, len2;
17 int dp[MAX][MAX];    //dp[i][j]：s1基因的前i个碱基和s2基因的前j个碱基中的最优匹配解
18 
19 map<char, int> gene;    //碱基和adjust下标一一映射
20 int adjust[5][5] = {    {5,-1,-2,-1,-3}, 
21                         {-1,5,-3,-2,-4},
22                         {-2,-3,5,-2,-2},
23                         {-1,-2,-2,5,-1},
24                         {-3,-4,-2,-1,0}};
25 
26 int main()
27 {
28     int T;
29     scanf("%d", &T);
30 
31     gene['A'] = 0; gene['C'] = 1;
32     gene['G'] = 2; gene['T'] = 3;
33     gene['-'] = 4;
34 
35     while (T--)
36     {
37         scanf("%d %s", &len1, s1);
38         scanf("%d %s", &len2, s2);
39 
40         memset(dp, 0, sizeof(dp));
41 
42         for (int i = 1; i <= len1; i++)
43             dp[i][0] = dp[i-1][0] + adjust[gene[s1[i - 1]]][gene['-']];    //上碱基各碱基仅匹配'-'时
44         for (int j = 1; j <= len2; j++)
45             dp[0][j] = dp[0][j-1] + adjust[gene['-']][gene[s2[j - 1]]];    //下碱基各碱基仅匹配'-'时
46 
47         for (int i = 1; i <= len1; i++)
48             for (int j = 1; j <= len2; j++)
49             {
50                 dp[i][j] = max(dp[i][j - 1] + adjust[gene['-']][gene[s2[j - 1]]],
51                     dp[i - 1][j] + adjust[gene[s1[i - 1]]][gene['-']]);        //上下碱基各匹配'-'时
52                 dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + adjust[gene[s1[i - 1]]][gene[s2[j - 1]]]);    //上下碱基相互配对时
53             }
54         printf("%d\n", dp[len1][len2]);
55     }
56 
57 
58     return 0;
59 }