解題思路：這道題最開始看錯題了，以為只能給長度小的那個字符串添加'-'，就往“編輯距離”這個模型去想。其實這里是兩個串都可以添加'-'，所以模型一下子就簡化了。這里可以用LCS的思路去做。

dp[i][j]表示兩個串分別是前i個和前j個的最大價值。

dp[i][j] = max(dp[i-1][j-1] + score[A[i]][B[j]],max(dp[i-1][j] + score[A[i]][-], dp[i][j-1] + score[-][B[j]]))，這里一定要注意不要犯這樣的錯誤：i和j的字符相等，就想當(dāng)然的認(rèn)為它是從dp[i-1][j-1]轉(zhuǎn)移過來的，實際上可能還會有別的匹配方式。這里老是犯錯，所以一定要記住。

另外這里初始化也是要注意的地方。

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std;const int maxn = 105; int n,m,dp[maxn][maxn],score[90][90]; char A[maxn],B[maxn];int main() {int t;char tmp[maxn];scanf("%d",&t);score['A']['A'] = score['C']['C'] = score['G']['G'] = score['T']['T'] = 5;score['A']['C'] = score['C']['A'] = -1;score['G']['A'] = score['A']['G'] = -2; score['G']['C'] = score['C']['G'] = -3;score['T']['A'] = score['A']['T'] = -1; score['T']['C'] = score['C']['T'] = -2; score['T']['G'] = score['G']['T'] = -2;score[' ']['A'] = score['A'][' '] = -3; score[' ']['C'] = score['C'][' '] = -4; score[' ']['T'] = score['T'][' '] = -1; score[' ']['G'] = score['G'][' '] = -2;while(t--){cin >> n;cin >> A+1;cin >> m;cin >> B+1;memset(dp,0,sizeof(dp));for(int i = 1; i <= n; i++)dp[i][0] = dp[i-1][0] + score[A[i]][' '];for(int i = 1; i <= m; i++)dp[0][i] = dp[0][i-1] + score[' '][B[i]];for(int i = 1; i <= n; i++)for(int j = 1; j <= m; j++){dp[i][j] = max(dp[i-1][j-1] + score[A[i]][B[j]],max(dp[i-1][j] + score[A[i]][' '], dp[i][j-1] + score[' '][B[j]]));}cout << dp[n][m] << endl;}return 0; }

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