2020 力扣杯！Code Your Future 春季全国编程大赛 - python 代码

第一题 拿硬币

LCP 06. 拿硬币

class Solution:
    def minCount(self, coins) -> int:
        cnt = 0
        for x in coins:
            cnt += x >> 1
            cnt += x & 1
        return cnt

第二题 传递信息

LCP 07. 传递信息

class Solution:
    def numWays(self, n: int, relation: List[List[int]], k: int) -> int:
        x = [0]
        m = {}
        for r in relation:
            if r[0] not in m:
                m[r[0]] = [r[1]]
            else:
                m[r[0]].append(r[1])
        for i in range(k):
            xx = []
            for c in x:
                if c in m:
                    xx.extend(m[c])
#             print(xx)
            x = xx
        cnt = 0
        for xxx in x:
            if xxx == n-1:
                cnt += 1
        return cnt

第三题 LCP 08. 剧情触发时间

LCP 08. 剧情触发时间

import bisect
class Solution:
    def getTriggerTime(self, increase: List[List[int]], requirements: List[List[int]]) -> List[int]:
        c = [0]
        r = [0]
        h = [0]
        
        for cc, rr, hh in increase:
            c.append(c[-1] + cc)
            r.append(r[-1] + rr)
            h.append(h[-1] + hh)
#         print(c)
#         print(r)
#         print(h)
        a = []
        for cc, rr, hh in requirements:
            ct = bisect.bisect_left(c, cc)
            rt = bisect.bisect_left(r, rr)
            ht = bisect.bisect_left(h, hh)
            xx = max([ct, rt, ht])
            if xx == len(c): xx = -1
            a.append(xx)
        return a

第四题 最小跳跃次数

LCP 09. 最小跳跃次数

class Solution:
    def minJump(self, jump: List[int]) -> int:
        l = len(jump)
        outi = []
        to = {}
        fr = {}
        ddd = [-1] * l
        for i in range(l):
            d = i + jump[i]
            if d >= l:
                outi.append(i)
            else:
                to[i] = d
                if d in fr:
                    fr[d].append(i)
                else:
                    fr[d] = [i]
        mi = min(outi)
        import queue
        q = queue.PriorityQueue()
        for oi in outi:
            ddd[oi] = 0
            q.put((0, oi))
#         print(q.queue)
#         print(fr)
        cm = l
        while True:
            i, n = q.get()
#             print(i, n)
            if n == 0:
                break
            for j in range(n+1, cm):
                if ddd[j] == -1 or ddd[j] > i+1:
#                     print('%d -> %d' %(n, j))
                    q.put((i+1, j))
                    ddd[j] = i+1
            cm = min(cm, n)
            
            if n in fr:
                for f in fr[n]:
#                     print(f,'f')
                    if ddd[f] == -1 or ddd[f] > i+1:
                        ddd[f] = i+1
#                         print('%d -> %d' %(n, f))
                        q.put((i+1, f))
        return i+1

第五题 二叉树任务调度

LCP 10. 二叉树任务调度
我自己还没做出来，参考了两个题解 littledva的题解 和 zqy1018的题解

两篇都是 dfs，但思路不太一样，第一篇是把每棵树的时间分为单核时间（就是不能并行的时间）和双核时间，答案是二者的和；第二篇是分别计算每棵子树最短时间和所有任务的合计时间。

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