大佬看看这个代码没问题吧，第二题 DFS，加了个简单的剪枝

import sys


def per(origin, target):
    for i in target:
        if i not in origin:
            return []
    res = []
    dfs(origin, target, [], "", res)
    return res


def dfs(old, target, path, now, res):
    if not old:
        # 遍历结束
        # 判断是否相等
        if now == target:
            res.append(path.copy())
        return
    if not is_in(now, target):
        # 未遍历结束
        # 判断是否已经不相等
        return
    cur = old[0]
    if cur not in target:
        path.append("d")
        dfs(old[1:], target, path, now, res)
        path.pop()
    else:
        path.append("d")
        dfs(old[1:], target, path, now, res)
        path.pop()

        path.append("l")
        dfs(old[1:], target, path, cur + now, res)
        path.pop()

        path.append("r")
        dfs(old[1:], target, path, now + cur, res)
        path.pop()


def is_in(ls1, ls2):
    # 判断 ls1 是否为 ls2 的连续子集
    str1 = "".join(map(str, ls1))
    str2 = "".join(map(str, ls2))
    return str1 in str2


if __name__ == "__main__":
    s = int(sys.stdin.readline().strip())
    ans = ""
    for _ in range(s):
        origin = sys.stdin.readline().strip()
        target = sys.stdin.readline().strip()
        ans += "{\n"
        res = per(origin, target)
        res.sort()
        # print(res)
        if res:
            for each_res in res:
                ans += " ".join(each_res) + " \n"
        ans += "}\n"
    if ans:
        ans = ans[:-1]
    print(ans)