题解

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

// 碱基相似度表
int sim[5][5] = {
    {5, -1, -2, -1, -3}, // A
    {-1, 5, -3, -2, -4}, // C
    {-2, -3, 5, -2, -2}, // G
    {-1, -2, -2, 5, -1}, // T
    {-3, -4, -2, -1, 0}  // -
};

// 将碱基转换为索引
int getIndex(char c) {
    switch (c) {
        case 'A': return 0;
        case 'C': return 1;
        case 'G': return 2;
        case 'T': return 3;
        case '-': return 4;
        default: return -1;
    }
}

int main() {
    int n, m;
    string s1, s2;
    cin >> n >> s1 >> m >> s2;

    // 动态规划数组
    vector<vector<int>> dp(n + 1, vector<int>(m + 1, 0));

    // 初始化
    for (int i = 1; i <= n; i++) {
        dp[i][0] = dp[i-1][0] + sim[getIndex(s1[i-1])][4];
    }
    for (int j = 1; j <= m; j++) {
        dp[0][j] = dp[0][j-1] + sim[4][getIndex(s2[j-1])];
    }

    // 动态规划
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            int match = dp[i-1][j-1] + sim[getIndex(s1[i-1])][getIndex(s2[j-1])];
            int insert1 = dp[i-1][j] + sim[getIndex(s1[i-1])][4];
            int insert2 = dp[i][j-1] + sim[4][getIndex(s2[j-1])];
            dp[i][j] = max({match, insert1, insert2});
        }
    }

    // 输出结果
    cout << dp[n][m] << endl;
    return 0;
}