poj 1080 Human Gene Functions

#include <iostream>            //DP
using namespace std;
int max(int a,int b,int c)
{
    int aa=a,bb=b,cc=c;
    if(aa<bb)swap(aa,bb);
    return aa>cc?aa:cc;
}
int matrix(char a,char b)
{
    if(a=='A')
    {
        switch (b)
        {
        case 'A':return 5;break;
        case 'C':return -1;break;
        case 'G':return -2;break;
        case 'T':return -1;break;
        default:return -3;break;
        }
    }
    else if(a=='C')
    {
        switch (b)
        {
        case 'A':return -1;break;
        case 'C':return 5;break;
        case 'G':return -3;break;
        case 'T':return -2;break;
        default:return -4;break;
        }
    }
    else if(a=='G')
    {
        switch (b)
        {
        case 'A':return -2;break;
        case 'C':return -3;break;
        case 'G':return 5;break;
        case 'T':return -2;break;
        default:return -2;break;
        }
    }
    else if(a=='T')
    {
        switch (b)
        {
        case 'A':return -1;break;
        case 'C':return -2;break;
        case 'G':return -2;break;
        case 'T':return 5;break;
        default:return -1;break;
        }
    }
    else
    {
        switch (b)
        {
        case 'A':return -3;break;
        case 'C':return -4;break;
        case 'G':return -2;break;
        default:return -1;break;
        }
    }
}    
char str1[110],str2[110];
int ans[110][110];
int dp(int i,int j)        //求字符串1,str1[0]-str1[i-1]与字符串2,str2[0]-str2[j-1] 所能匹配的最大分数
{
    if(ans[i][j]!=0)
        return ans[i][j];
    else if(i==0&&j==0)
            ans[i][j]=0;
    else if(i==0)
        ans[i][j]=dp(i,j-1)+matrix('-',str2[j-1]);
    else if(j==0)
        ans[i][j]=dp(i-1,j)+matrix(str1[i-1],'-');
    else
    {
        ans[i][j]=max(dp(i-1,j-1)+matrix(str1[i-1],str2[j-1]);
        dp(i-1,j)+matrix(str1[i-1],'-'),dp(i,j-1)+matrix('-',str2[j-1]));
    }
    return ans[i][j];
}
int main()
{
    int t,i,len1,len2;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%s",&len1,str1);
        scanf("%d%s",&len2,str2);
        memset(ans,0,sizeof(ans));
        printf("%d\n",dp(len1,len2));
    }
    return 0;
}