【程设】Python的三道小题

1.热血格斗场

    该题需要用有序队列实现，用c++中的map，set都可以轻松实现。但是由于python的dict是无序表，因此采用对list容器进行二分查找的方法解决该题。

from math import *
def minsert(newone):
    global mydict
    p = 0
    q = len(mydict) - 1
    m = floor((p+q)/2)
    while(p != q):
        if mydict[m][0] < newone[0]:
            p = m + 1
        else:
            q = m
        m = floor((p+q)/2)
    mydict.insert(p, newone)
    return p
n = int(input())
mydict = [[1e9, 1]]
for i in range(n):
    newone = list(map(int, input().split()))
    newone.reverse()
    j = minsert(newone)
    if j == 0:
        print(newone[1], mydict[j + 1][1])
    elif j == len(mydict) - 1:
        print(newone[1], mydict[j - 1][1])
    else:
        if (mydict[j + 1][0] + mydict[j - 1][0] < 2 * newone[0]):
            print(newone[1], mydict[j + 1][1])
        else:
            print(newone[1], mydict[j - 1][1])

2.拯救行动

    一道深度搜索跑迷宫的题目

def inmap(i, j):
    global m
    global n
    global mydict
    if i >= 0 and i <= m - 1 and j >= 0 and j <= n - 1 and (
            mydict[i][j] == '@' or mydict[i][j] == 'x' or mydict[i][j] == 'a'):
        return True
    else:
        return False


S = int(input())
for d in range(S):
    m, n = map(int, input().split())
    #print(m, n)
    mydict = []
    myqueue = [[], []]
    myso = {}
    for i in range(m):
        mydict.append(input())
    # print(mydict)
    for i in range(m):
        for j in range(n):
            if mydict[i][j] == 'r':
                warrior = (i, j)
            elif mydict[i][j] == 'a':
                princess = (i, j)
            elif mydict[i][j] == 'x':
                myso[(i, j)] = -1
    myqueue[0].append(warrior)
    cnt = 0
    flag = False
    while len(myqueue[0]):
        cnt = cnt + 1
        #print(cnt)
        for v in myqueue[0]:
            #print(v)
            if v == princess:
                print(cnt - 1)
                flag = True
                break
            #print(v in myso.keys())
            if (v in myso.keys()):
                # print(v)
                if myso[v] == -1:
                    myso[v] = cnt
                    myqueue[1].append(v)
                    continue
                elif myso[v] == 0 or myso[v] == cnt:
                    continue
            myso[v] = 0
            if inmap(v[0] - 1, v[1]):
                myqueue[1].append((v[0] - 1, v[1]))
            if inmap(v[0] + 1, v[1]):
                myqueue[1].append((v[0] + 1, v[1]))
            if inmap(v[0], v[1] - 1):
                myqueue[1].append((v[0], v[1] - 1))
            if inmap(v[0], v[1] + 1):
                myqueue[1].append((v[0], v[1] + 1))
        if(flag):
            break
        del myqueue[0]
        myqueue.append([])
    if flag == False:
        print('Impossible')

三.拨钟问题

    实际上是解一个矩阵方程的问题。

    直接使用枚举法超时了，于是只遍历操作1，2，3，其他的均可有1，2，3确定。

from copy import deepcopy
def move_clock(k, n = 1):
    global temp
    global move
    global op
    for i in range(9):
        temp[i] = (temp[i] + n * move[k][i]) % 4

def find(depth):
    global temp
    global min_num
    global min_op
    global op
    if depth == 9:
        if sum(temp) == 0 and sum(op) < min_num:
            min_num = sum(op)
            min_op = deepcopy(op)
        return
    elif depth < 3:
        for i in range(4):
            move_clock(depth)
            op[depth] = (op[depth] + 1) % 4
            find(depth + 1)
        return
    elif depth < 7:
        op[depth] = (4 - temp[depth - 3]) % 4
    elif depth == 7:
        op[depth] = (4 - temp[6]) % 4
    else:
        op[depth] = (4 - temp[4]) % 4
    move_clock(depth, op[depth])
    find(depth + 1)
    move_clock(depth, -op[depth])
    op[depth] = 0
    return
op = [0 for i in range(9)]
temp = [] #temporary result
min_num = 28
min_op = [0 for i in range(9)]
move = [[0 for i in range(9)] for j in range(9)]
move[0][0] = move[0][1] = move[0][3] = move[0][4] = 1
move[1][0] = move[1][1] = move[1][2] = 1
move[2][1] = move[2][2] = move[2][4] = move[2][5] = 1
move[3][0] = move[3][3] = move[3][6] = 1
move[4][1] = move[4][3] = move[4][4] = move[4][5] = move[4][7] = 1
move[5][2] = move[5][5] = move[5][8] = 1
move[6][3] = move[6][4] = move[6][6] = move[6][7] = 1
move[7][6] = move[7][7] = move[7][8] = 1
move[8][4] = move[8][5] = move[8][7] = move[8][8] = 1
for i in range(3):
    temp += input().split()
temp = list(map(int, temp))
find(0)
for i in range(9):
    for j in range(min_op[i]):
        print(i + 1,end = ' ')