span_layout_search.py

"""
span_layout_search.py
跨径布置确定性搜索：给定桥梁总长和跨数范围，在"等跨"与"边跨缩短"两类
经典布置方案之间搜索，复用 design_search.find_adequate_reinforcement
判断每个候选跨径组合是否结构可行，在可行方案里选出混凝土用量最省的一个。

与 design_search.py 一样全部本地计算，不涉及 LLM。
"""
from dataclasses import dataclass
from typing import List, Optional, Tuple

import design_search
from export_utils import calculate_bridge_quantities

END_SPAN_RATIO = 0.8  # 边跨/中跨长度比：连续梁桥边跨适当缩短是常见工程做法，
                      # 可以降低边跨的正弯矩，让跨径分布更合理


@dataclass
class SpanLayoutCandidate:
    spans: List[float]
    pattern: str  # "等跨" 或 "边跨缩短"
    beam_h: float
    num_girders: int
    bar_diameter_mm: int
    worst_utilization: float
    total_concrete_m3: float


@dataclass
class SpanLayoutResult:
    found: bool
    best: Optional[SpanLayoutCandidate]
    candidates: List[SpanLayoutCandidate]  # 所有可行方案，按混凝土用量升序排列
    message: str


def _generate_span_patterns(total_length: float, num_spans: int) -> List[Tuple[str, List[float]]]:
    """返回 (方案名, 跨径列表) 列表：等跨方案总是有；跨数 >= 3 时额外生成边跨缩短方案"""
    equal_span = total_length / num_spans
    patterns = [("等跨", [round(equal_span, 2)] * num_spans)]

    if num_spans >= 3:
        # 2 * (k*x) + (num_spans-2) * x = total_length，解出中跨长度 x
        k = END_SPAN_RATIO
        x = total_length / (2 * k + (num_spans - 2))
        end_span = round(k * x, 2)
        mid_span = round(x, 2)
        spans = [end_span] + [mid_span] * (num_spans - 2) + [end_span]
        patterns.append(("边跨缩短", spans))

    return patterns


def find_span_layout(
    total_length: float,
    bridge_w: float,
    pier_h: float,
    q_live: float,
    num_spans_min: int = 2,
    num_spans_max: int = 5,
    beam_h_start: float = 2.0,
    current_num_girders: int = 5,
    current_bar_diameter_mm: int = 25,
) -> SpanLayoutResult:
    """
    在 [num_spans_min, num_spans_max] 跨数范围内，对"等跨"和"边跨缩短"两类
    布置方案分别做结构可行性搜索（复用 design_search 的梁高/梁片数/主筋
    直径搜索），返回所有可行方案（按混凝土用量升序），并推荐用量最省的一个。
    """
    candidates: List[SpanLayoutCandidate] = []

    for num_spans in range(num_spans_min, num_spans_max + 1):
        for pattern_name, spans in _generate_span_patterns(total_length, num_spans):
            search_result = design_search.find_adequate_reinforcement(
                spans=spans,
                beam_h=beam_h_start,
                pier_h=pier_h,
                bridge_w=bridge_w,
                q_live=q_live,
                current_num_girders=current_num_girders,
                current_bar_diameter_mm=current_bar_diameter_mm,
            )
            if not search_result.found:
                continue

            quantities = calculate_bridge_quantities(
                spans=spans, beam_h=search_result.beam_h, pier_h=pier_h, bridge_width=bridge_w
            )
            candidates.append(
                SpanLayoutCandidate(
                    spans=spans,
                    pattern=pattern_name,
                    beam_h=search_result.beam_h,
                    num_girders=search_result.num_girders,
                    bar_diameter_mm=search_result.bar_diameter_mm,
                    worst_utilization=search_result.worst_utilization,
                    total_concrete_m3=quantities["total_concrete_m3"],
                )
            )

    if not candidates:
        message = (
            f"总长 {total_length:.1f}m 在 {num_spans_min}~{num_spans_max} 跨范围内，"
            "尝试的跨径布置方案都无法找到满足承载力要求的截面/配筋方案。"
            "建议扩大跨数搜索范围（增加跨数、缩短单跨），或考虑增大桥面宽度/减小总长。"
        )
        return SpanLayoutResult(found=False, best=None, candidates=[], message=message)

    candidates.sort(key=lambda c: c.total_concrete_m3)
    best = candidates[0]
    message = (
        f"已对总长 {total_length:.1f}m 在 {num_spans_min}~{num_spans_max} 跨范围内搜索，"
        f"找到 {len(candidates)} 个满足承载力要求的可行方案。推荐方案："
        f"{len(best.spans)}跨（{best.pattern}）{best.spans}，梁高 {best.beam_h:.2f}m，"
        f"梁片数 {best.num_girders}，主筋直径 {best.bar_diameter_mm}mm，"
        f"最大利用率 {best.worst_utilization:.2f}，混凝土总用量约 "
        f"{best.total_concrete_m3:.1f}m³（可行方案中最省）。"
    )
    return SpanLayoutResult(found=True, best=best, candidates=candidates, message=message)