POJ 1080 Human Gene Functions(DP)

题目链接

这个匹配矩阵，很有特点，鼓励相同，鼓励匹配，不鼓励匹配空。状态转移很类似于LCS，初始化做好，OK了。

 1 #include <cstring>
 2 #include <string>
 3 #include <cstdio>
 4 #include <iostream>
 5 using namespace std;
 6 int c[5][5] = {{5,-1,-2,-1,-3},{-1,5,-3,-2,-4},{-2,-3,5,-2,-2},{-1,-2,-2,5,-1},{-3,-4,-2,-1,0}};
 7 char s1[101],s2[101];
 8 int dp[101][101];
 9 int judge(char a)
10 {
11     if(a == 'A')
12     return 0;
13     else if(a == 'C')
14     return 1;
15     else if(a == 'G')
16     return 2;
17     else if(a == 'T')
18     return 3;
19     return 0;
20 }
21 int main()
22 {
23     int t,len1,len2,i,j,x,y,sum;
24     scanf("%d",&t);
25     while(t--)
26     {
27         memset(dp,0,sizeof(dp));
28         scanf("%d%s%d%s",&len1,s1,&len2,s2);
29         sum = 0;
30         for(i = 1;i <= len1;i ++)
31         {
32             x = judge(s1[i-1]);
33             y = judge(s2[0]);
34             dp[i][1] = sum + c[x][y];
35             dp[i][1] = max(dp[i-1][1]+c[x][4],dp[i][1]);
36             sum += c[x][4];
37         }
38         sum = 0;
39         for(i = 1;i <= len2;i ++)
40         {
41             x = judge(s1[0]);
42             y = judge(s2[i-1]);
43             dp[1][i] = sum + c[x][y];
44             dp[1][i] = max(dp[1][i-1]+c[4][y],dp[1][i]);
45             sum += c[4][y];
46         }
47         for(i = 2; i <= len1;i ++)
48         {
49             for(j = 2;j <= len2;j ++)
50             {
51                 x = judge(s1[i-1]);
52                 y = judge(s2[j-1]);
53                 dp[i][j] = dp[i-1][j-1] + c[x][y];
54                 dp[i][j] = max(dp[i-1][j]+c[x][4],dp[i][j]);
55                 dp[i][j] = max(dp[i][j-1]+c[4][y],dp[i][j]);
56             }
57         }
58         printf("%d\n",dp[len1][len2]);
59     }
60     return 0;
61 }

 

posted @ 2013-01-22 10:11  Naix_x  阅读(166)  评论(0)    收藏  举报