"""Reward Calculator — 计算一条审批轨迹的奖励分数。

诚实分级设计 (对齐兴邦方案):
  Level 1 (现在能做,不需要用户交互):
    - 任务完成 +10
    - 工具调用成功率
    - 重复调用惩罚
    - 时间惩罚
  Level 2 (需前端加 👍/👎 按钮):
    - 用户显式反馈
  Level 3 (需人工标注):
    - 专家打分

本文件实现 Level 1 + 预留 Level 2/3 参数。
不做不诚实的信号: 拒绝用"还款表现"当奖励(周期1-3年无法闭环)。
"""

from __future__ import annotations
from typing import Any


def compute_reward_v1(
    trajectory: dict[str, Any],
    feedback: str | None = None,
    expert_score: float | None = None,
) -> float:
    """计算一条轨迹的奖励(Leve1 l 内置指标 + 可选 Level2 反馈 + Level3 专家分)。

    Args:
        trajectory: 轨迹 dict(含 steps 和 final_result)
        feedback: Level2, "positive"/"negative"
        expert_score: Level3, 1-5 分
    Returns:
        float 奖励分
    """
    steps = trajectory.get("steps", [])
    final = trajectory.get("final_result", {})
    if not steps:
        return 0.0

    reward = 0.0

    # 1. 任务完成奖励
    if final.get("success"):
        reward += 10.0
    else:
        reward -= 10.0

    # 2. 工具调用成功率(成功工具 / 总工具)
    total_tool_calls = 0
    tool_success = 0
    for s in steps:
        tools = s.get("tools", [])
        total_tool_calls += len(tools)
        if tools:
            tool_success += 1  # 该步骤至少调了工具视为尽力(简化)
    if total_tool_calls > 0:
        reward += (tool_success / max(1, total_tool_calls)) * 5.0

    # 3. 步骤数惩罚(越少越好,常规 5 步流水线 5-15 步为佳)
    n_steps = len(steps)
    optimal = 5
    step_penalty = abs(n_steps - optimal) * 0.3
    reward -= step_penalty

    # 4. 时间惩罚(超过 600 秒 = 10 分钟开始扣)
    duration_ms = final.get("duration_ms", 0)
    if duration_ms > 600_000:
        reward -= ((duration_ms - 600_000) / 60_000) * 1.0

    # ── Level 2: 用户反馈 (可选) ──
    if feedback == "positive":
        reward += 20.0
    elif feedback == "negative":
        reward -= 10.0

    # ── Level 3: 专家打分 (可选, 1-5 归一化到 -10~+10) ──
    if expert_score is not None:
        reward += (expert_score - 3) * 5.0

    reward = max(-30.0, min(reward, 50.0))  # 裁剪
    return round(reward, 2)