题意读了半年，唉，给你两串字符，然后长度不同，你可以用‘-’把它们补成相同长度，补在哪里取决于得分，它会给你一个得分表，问你最大得分

跟LCS很像的DP数组 dp[i][j]表示第一个字符串取第i个元素第二个字符串取第三个元素，然后再预处理一个得分表加上即可

得分表：

score['A']['A'] = score['C']['C'] = score['G']['G'] = score['T']['T'] = 5;

	score['A']['C'] = score['C']['A'] = -1;
	score['A']['G'] = score['G']['A'] = -2;
	score['A']['T'] = score['T']['A'] = -1;
	score['A']['-'] = score['-']['A'] = -3;
	score['C']['G'] = score['G']['C'] = -3;
	score['C']['T'] = score['T']['C'] = -2;
	score['C']['-'] = score['-']['C'] = -4;
	score['G']['T'] = score['T']['G'] = -2;
	score['G']['-'] = score['-']['G'] = -2;
	score['T']['-'] = score['-']['T'] = -1;

	score['-']['-'] = -inf;

dp[i][j] = ：

dp[i-1][j] + score[s1[i-1]][‘-‘]或者

dp[i][j-1] + score[‘-‘][s2[j-1]]或者

dp[i-1][j-1] + score[s1[i-1]][s2[j-1]]或者

这三者之中取最大的

然后就是边界问题我给忘记了

不够细心，若单单是i==0或者j==0，边界问题就出现了，边界不可能为0的，所以还得处理一下边界

#include<iostream>
#include<cstdio>
#include<list>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<cmath>
#include<memory.h>
#include<set>
#include<cctype>

#define ll long long

#define LL __int64

#define eps 1e-8

#define inf 0xfffffff

//const LL INF = 1LL<<61;

using namespace std;

//vector<pair<int,int> > G;
//typedef pair<int,int > P;
//vector<pair<int,int> > ::iterator iter;
//
//map<ll,int >mp;
//map<ll,int >::iterator p;

const int N = 1000 + 5;

int dp[N][N];
int score[200][200];

void init() {
	score['A']['A'] = score['C']['C'] = score['G']['G'] = score['T']['T'] = 5;

	score['A']['C'] = score['C']['A'] = -1;
	score['A']['G'] = score['G']['A'] = -2;
	score['A']['T'] = score['T']['A'] = -1;
	score['A']['-'] = score['-']['A'] = -3;
	score['C']['G'] = score['G']['C'] = -3;
	score['C']['T'] = score['T']['C'] = -2;
	score['C']['-'] = score['-']['C'] = -4;
	score['G']['T'] = score['T']['G'] = -2;
	score['G']['-'] = score['-']['G'] = -2;
	score['T']['-'] = score['-']['T'] = -1;

	score['-']['-'] = -inf;
}

int main () {
	init();
	int t;
	char s1[N];
	char s2[N];
	scanf("%d",&t);
	while(t--) {
		int n,m;
		memset(dp,0,sizeof(dp));
		scanf("%d %s",&n,s1);
		scanf("%d %s",&m,s2);
		for(int i=1;i<=n;i++)
			dp[i][0] = dp[i-1][0] + score[s1[i-1]]['-'];//边界处理
		for(int j=1;j<=m;j++)
			dp[0][j] = dp[0][j-1] + score['-'][s2[j-1]];//边界处理
		for(int i=1;i<=n;i++) {
			for(int j=1;j<=m;j++) {
				int t1 = dp[i-1][j] + score[s1[i-1]]['-'];
				int t2 = dp[i][j-1] + score['-'][s2[j-1]];
				int t3 = dp[i-1][j-1] + score[s1[i-1]][s2[j-1]];
				int maxn = max(t1,t2);
				dp[i][j] = max(maxn,t3);
			}
		}
		printf("%d\n",dp[n][m]);
	}
	return 0;
}