一，搜索连通域 

M, N = list(map(int, input().split(',')))
print(M,N)
book = []
for i in range(M):line = list(map(int, input().split(',')))book.append(line)
print(book)
# M=N=3
# book=[[1, 0, 0], [0, 1, 0], [0, 1, 1]]
print('book=',book)
class Solution:def __init__(self,pos):self.pos=pos#给定的矩阵print('self.pos',self.pos)print(self.pos[0][0])self.dp=[]#保存所有区域的人数self.cnt=0#记录当前区域的人数#搜索联通区域def dfs(self,i,j):if 0<=i<M and 0<=j<N:if self.pos[i][j]==1:#确保递归遍历的时候有人才加人self.cnt+=1self.pos[i][j] =0#遍历过置0,避免重复搜索self.dfs(i, j-1)self.dfs(i, j+1)self.dfs(i-1, j)self.dfs(i+1, j)self.dfs(i - 1, j-1)self.dfs(i - 1, j +1)self.dfs(i +1, j - 1)self.dfs(i +1, j +1)def solver(self):for i in range(M):for j in range(N):if self.pos[i][j]==1:self.cnt=0#清零人数重新计数self.dfs(i,j)if self.cnt>0:self.dp.append(self.cnt)return self.dp
s=Solution(book)
dp=s.solver()
print('球队个数={},球队的最大群体人数={}'.format(len(dp),max(dp)))

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二，简单二叉树的实现

#简单二叉树的实现
class Node:def __init__(self,number):self.number=numberself.lchild=Noneself.rchild=None
class Tree:lis=[]def __init__(self):self.root=None#增加节点def add(self,number):node=Node(number)if self.root==None:self.root=nodeprint('add root')Tree.lis.append(self.root)print('Tree.lis=', Tree.lis)else:while True:point=Tree.lis[0]print('point=',point)if point.lchild==None:point.lchild=nodeTree.lis.append(point.lchild)print('add lchild')print('Tree.lis=', Tree.lis)returnif point.rchild==None:point.rchild=nodeTree.lis.append(point.rchild)print('add rchild')print('Tree.lis=', Tree.lis)Tree.lis.pop(0)#更换root节点print('Tree.lis=', Tree.lis)return
if __name__ == '__main__':t=Tree()L=[1,2,3,4,5,6,7]for x in L:t.add(x)# print('success')

三，找列表里的最长字符串

#找列表里的最长字符串
res=['a','v','f','sd','456','dfgg']
num_list=[len(i) for i in res]
max_length_str=res[num_list.index(max(num_list))]
print(max_length_str)#找列表里的最长字符串
a=sorted(res,key=lambda x:len(x),reverse=True)
print(a[0])

四，找两个字符串的最长公共字符串(暴力解法)

def function(s1):#找所有子字符串的结果a=[]for i in range(0,len(s1)):for j in range(i+1,len(s1)+1):a.append(s1[i:j])return as1='sadsadarefcsfdsfds'
s2='sadsafcvwertdfgdfg'
res1=function(s1)
res2=function(s2)res=[]
for i in res1:if i in res2:#找共同字符串res.append(i)
#找出共同字符串中的最长字符串
res=sorted(res,key=lambda x:len(x),reverse=True)
print(res[0])

五，如题

A=list(map(int,input().split(',')))
print(A)# A=[1,3,4,4]
for i in range(len(A)-1):if sum(A[:i])==A[i] and A[i]==sum(A[i+1:]):print(A[i])else:print(False)

六，如题

a1,a2=input().split('-')
print('a1,a2=',a1,a2)
a1=list(map(int,a1.split(',')))
a2,k=a2.split(':')a2=list(map(int,a2.split(',')))
k=int(k)
print('a1,a2,k=',a1,a2,k)tem=[]
for i in a1:tem+=[i+j for j in a2]
print('tem=',tem)
print(list(map(str,sorted(tem,reverse=True)))[:k])
print(','.join(list(map(str,sorted(tem,reverse=True)))[:k]))

七，如题

X,Y,Z=sorted(list(map(int,input().split(' '))))
print(X,Y,Z)
def solve(X,Y,Z):cnt = 0for a in range(1,X+1):for b in range(1,Y+1):#最短边min_c=abs(a-b)+1# print(min_c)# 最长边max_c=min(a+b-1,Z)# print(max_c)cnt+=max_c-min_c+1cnt%=1000000007return cnt
cnt=solve(X,Y,Z)
print('cnt=',cnt)

 

八，如题，采用贪心算法

N=int(input())
line=[]
for i in range(N):a,b=sorted(list(map(int,input().split(' '))))line.append([a,b])
print(line)# line=[[3, 6], [1, 3], [2, 5]]
line=sorted(line,key=lambda x:x[1])
print('line=',line)ret = [line[0]]
print('ret=',ret)
for item in line[1:]:print('item=',item)if ret[-1][1]>item[0]:passelse:ret.append(item)
print(ret)
print(len(ret))

九，如题，

N,M,P=list(map(int,input().split(' ')))
print('N={},M={},p={}'.format(N,M,P))
A=list(map(int,input().split(' ')))
print(A)# N,M,P=3,4,2
# A=[5,3,1]
for i in range(M):flag,t=input().split(' ')t=int(t)if flag=='A':A[t-1]+=1else:A[t - 1] -= 1print('A=',A)
# A=[4,5,2]
A=list(zip(A,range(1,len(A)+1)))
print('A=',A)
#按照份数进行排序
A=sorted(A,reverse=True)
print('A=',A)
#排名第一的第二种食物，value=1
ret={A[0][1]:1}
print('ret=',ret)
for i in range(1,len(A)):if A[i][0]==A[i-1][0]:ret[A[i][1]]=ret[A[i][0]]else:#key->valueret[A[i][1]] =i+1
print(ret)
print(ret[P])

十.教学任务指派问题,匈牙利算法,由谁去做某某件事情总体的时间或者效率最高

#教学任务指派问题
#匈牙利算法,由谁去做某某件事情最合适import numpy as np
from scipy.optimize import linear_sum_assignmentdef printf(row_ind,col_ind):#输出print("最优教师课程指派：")for i in range(len(row_ind)):print("教师",row_ind[i], "->课程", col_ind[i],end='; ')print()
#教师与课程一样多
#各个教师对各个课的擅长程度矩阵
goodAt =np.array([[18, 5, 7, 16],[10, 16, 6, 5],[11, 6, 4, 7],[13, 12, 9, 11]])
print('==每个老师对课程的擅长矩阵:\n', goodAt)
#由最大值减去矩阵,变成求解最小值问题
weakAt = 20 - goodAt
row_ind, col_ind = linear_sum_assignment(weakAt)
print('==row_ind:', row_ind)#开销矩阵对应的行索引
print('==col_ind:', col_ind)#对应行索引的最优指派的列索引
print('==goodAt[row_ind, col_ind]:', goodAt[row_ind, col_ind])#提取每个行索引的最优指派列索引所在的元素，形成数组
print('最优结果为', goodAt[row_ind, col_ind].sum())#数组求和
printf(row_ind, col_ind)