rt，用的是 wei_xin 大佬在这个贴 发的思路：

#include <iostream>

using namespace std;

const int N = 5010;

int T;
int n, x, y; 
int f[N][N];
int k[N];

int cost(int l, int r) {
    return min((k[r] - k[l]) * x, y);
}

int main() {
    cin >> T;
    while (T--) {
        cin >> n >> x >> y;
        string a, b; getline(cin, a);
        getline(cin, a), getline(cin, b);
        int cnt = 0;
        for (int i = 0; i < n; i++) 
            if (a[i] != b[i]) k[++cnt] = i;
        if (cnt % 2) {
            cout << -1 << endl;
            continue;
        }
        if (cnt == 2) {
            bool flag = false;
            for (int i = 0; i < n - 1; i++)
                if (a[i] != b[i] && a[i + 1] != b[i + 1]) {
                    flag = true;
                    break;
                }
            if (flag) {
                if (n == 2) cout << x << endl;
                else cout << min(x, 2 * y) << endl;
                continue;
            }
            cout << min(y, (k[2] - k[1]) * x) << endl;
            continue;
        }//cnt = 2 时候的特判
        for (int i = 1; i < cnt; i++)
            for (int j = i + 1; j <= cnt; j += 2) 
                if (k[i] + 1 != k[j]) f[i][j] = min((k[j] - k[i]) * x, y);
                else f[i][j] = min(x, y * 2);//初始化
        for (int len = 4; len <= n; len++) {
            for (int l = 1; l < n; l++) {
                int r = l + len - 1;
                if (r > n) break;
                f[l][r] = min(min(f[l + 2][r] + cost(l, l + 1), f[l][r - 2] + cost(r - 1, r)), f[l + 1][r - 1] + y/*cost(l, r)*/);//dp
            }
        }
        cout << f[1][cnt] << endl;
    }
    return 0;
}