2025-06-23：破解锁的最少时间Ⅰ。用go语言，Bob 被困在一个地窖，必须解开 n 把锁才能逃出。每把锁需要一定的能量才

package main

import (
    "fmt"
    "math"
)

func findMinimumTime(strength []int, k int)int {
    n := len(strength)
    S := n * 2
    T := S + 1

    // rid 为反向边在邻接表中的下标
    type neighbor struct{ to, rid, cap, cost int }
    g := make([][]neighbor, T+1)
    addEdge := func(from, to, cap, cost int) {
        g[from] = append(g[from], neighbor{to, len(g[to]), cap, cost})
        g[to] = append(g[to], neighbor{from, len(g[from]) - 1, 0, -cost})
    }
    for i, s := range strength {
        // 枚举这个锁是第几次开的
        for j := range n {
            x := 1 + k*j
            addEdge(i, n+j, 1, (s-1)/x+1)
        }
        addEdge(S, i, 1, 0)
    }
    for i := n; i < n*2; i++ {
        addEdge(i, T, 1, 0)
    }

    // 下面是最小费用最大流模板
    dis := make([]int, len(g))
    type vi struct{ v, i int }
    fa := make([]vi, len(g))
    inQ := make([]bool, len(g))
    spfa := func()bool {
        for i := range dis {
            dis[i] = math.MaxInt
        }
        dis[S] = 0
        inQ[S] = true
        q := []int{S}
        forlen(q) > 0 {
            v := q[0]
            q = q[1:]
            inQ[v] = false
            for i, e := range g[v] {
                if e.cap == 0 {
                    continue
                }
                w := e.to
                newD := dis[v] + e.cost
                if newD < dis[w] {
                    dis[w] = newD
                    fa[w] = vi{v, i}
                    if !inQ[w] {
                        inQ[w] = true
                        q = append(q, w)
                    }
                }
            }
        }
        // 循环结束后所有 inQ[v] 都为 false，无需重置
        return dis[T] < math.MaxInt
    }

    minCost := 0
    for spfa() {
        // 沿 st-end 的最短路尽量增广
        // 特别地，如果建图时所有边的容量都设为 1，那么 minF 必然为 1，下面第一个 for 循环可以省略
        minF := math.MaxInt
        for v := T; v != S; {
            p := fa[v]
            minF = min(minF, g[p.v][p.i].cap)
            v = p.v
        }
        for v := T; v != S; {
            p := fa[v]
            e := &g[p.v][p.i]
            e.cap -= minF
            g[v][e.rid].cap += minF
            v = p.v
        }
        minCost += dis[T] * minF
    }
    return minCost
}

func main() {
    strength := []int{3, 4, 1}
    K := 1
    fmt.Println(findMinimumTime(strength, K))
}

# -*-coding:utf-8-*-

from collections import deque
import math

def find_minimum_time(strength, K):
    n = len(strength)
    S = n * 2
    T = S + 1

    # 邻接表中每条边用字典表示
    # neighbor: to, rid (反向边下标), cap, cost
    g = [[] for _ in range(T + 1)]

    def add_edge(frm, to, cap, cost):
        g[frm].append({'to': to, 'rid': len(g[to]), 'cap': cap, 'cost': cost})
        g[to].append({'to': frm, 'rid': len(g[frm]) - 1, 'cap': 0, 'cost': -cost})

    for i, s in enumerate(strength):
        # 枚举这个锁是第几次开的
        for j in range(n):
            x = 1 + K * j
            # 计算开锁需要的时间 (向上取整用 (s-1)//x + 1)
            cost = (s - 1) // x + 1
            add_edge(i, n + j, 1, cost)
        add_edge(S, i, 1, 0)

    for i in range(n, n * 2):
        add_edge(i, T, 1, 0)

    dis = [math.inf] * (T + 1)
    fa = [None] * (T + 1)  # 存父节点 (v, i)
    in_q = [False] * (T + 1)

    def spfa():
        nonlocal dis, fa, in_q
        for i in range(len(dis)):
            dis[i] = math.inf
            fa[i] = None
            in_q[i] = False

        dis[S] = 0
        q = deque([S])
        in_q[S] = True

        while q:
            v = q.popleft()
            in_q[v] = False
            for i, e in enumerate(g[v]):
                if e['cap'] == 0:
                    continue
                w = e['to']
                new_dist = dis[v] + e['cost']
                if new_dist < dis[w]:
                    dis[w] = new_dist
                    fa[w] = (v, i)
                    if not in_q[w]:
                        in_q[w] = True
                        q.append(w)
        return dis[T] < math.inf

    min_cost = 0
    while spfa():
        # 找到增广路径的最小容量
        min_f = math.inf
        v = T
        while v != S:
            pv, pi = fa[v]
            min_f = min(min_f, g[pv][pi]['cap'])
            v = pv

        # 沿路径更新容量
        v = T
        while v != S:
            pv, pi = fa[v]
            g[pv][pi]['cap'] -= min_f
            rid = g[pv][pi]['rid']
            g[v][rid]['cap'] += min_f
            v = pv

        min_cost += dis[T] * min_f

    return min_cost


if __name__ == "__main__":
    strength = [3, 4, 1]
    K = 1
    print(find_minimum_time(strength, K))