dp黑洞再次憑一己之力寫對了一道dp水題，這個題很簡單，跟最長公共子序列一樣。

#include<iostream>
#include<cstdio>
#include<string.h>
#include<algorithm>
#include<cmath>
using namespace std;

int t,f[105][105],len1,len2;
char s1[105],s2[105];

int get(char ch)
{
    if (ch=='A') return 1;
    if (ch=='C') return 2;
    if (ch=='G') return 3;
    if (ch=='T') return 4;
    return 5;
}

int find(char a,char b)
{
    int x=get(a);
    int y=get(b);
    if (x==y) return 5;
    else if (x*y==2 || x*y==4 || x*y==20) return -1;
    else if (x*y==3 || x*y==8 || x*y==12 || x*y==15) return -2;
    else if (x*y==5 || x*y==6) return -3;
    else if (x*y==10) return -4;
}

int main()
{
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d",&len1);
        for (int i=1; i<=len1; i++)
        {
            cin>>s1[i];
        }
        scanf("%d",&len2);
        for (int i=1; i<=len2; i++)
        {
            cin>>s2[i];
        }
        memset(f,0,sizeof(f));
        for (int i=1; i<=len1; i++)
        {
            f[i][0]=f[i-1][0]+find('-',s1[i]);
        }
        for (int i=1; i<=len2; i++)
        {
            f[0][i]=f[0][i-1]+find('-',s2[i]);
        }
        for (int i=1; i<=len1; i++)
        {
            for (int j=1; j<=len2; j++)
            {
                f[i][j]=max(f[i-1][j]+find(s1[i],'-'),f[i][j-1]+find(s2[j],'-'));
                f[i][j]=max(f[i][j],f[i-1][j-1]+find(s1[i],s2[j]));
                //cout<<"i="<<i<<" j="<<j<<endl;
                //cout<<"f1="<<f[i-1][j]+find(s1[i],'-')<<" f2="<<f[i][j-1]+find(s2[j],'-')<<" f3="<<f[i-1][j-1]+find(s1[i],s2[j])<<endl;
            }
        }
        printf("%d\n",f[len1][len2]);
    }
    return 0;
}