蓝桥杯 并查集汇总学习 及其代码
这里记录一下在刷蓝桥杯的并查集的题时写的一些代码
蓝桥幼儿园
蓝桥幼儿园 :https://www.lanqiao.cn/problems/1135/learning/
# https://www.lanqiao.cn/problems/1135/learning/
N,M = map(int,input().split())
f = [i for i in range(N+1)]
def find(x):
if f[x] != x:
f[x] = find(f[x]) # 路径压缩
return f[x]
# 合并两个
def union(x,y):
fx = find(x)
fy = find(y)
if fx != fy:
f[fy] = f[fy]
for i in range(M):
op,x,y = map(int,input().split())
if op == 1:
union(x,y)
elif op == 2:
if find(f[x]) == find(f[y]):
print('YES')
else:
print('NO')
七段码
七段码:https://www.lanqiao.cn/problems/595/learning/
# https://www.lanqiao.cn/problems/595/learning/
import itertools
class UnionFindSet():
def __init__(self,n):
self.count = 0 # 当前连通分量数目
# self.count = n # 不连通区域
self.father=[x for x in range(n)]
def find(self,x):
root = x
while self.father[root]!=root: # 找根节点
root=self.father[root]
# 路径压缩
while x != root:
o = self.father[x] # 找x的父节点
self.father[x] = root # 把x的父节点设置成刚才找到的根
x = o # 往上一层
return root
def union(self,x,y):
x,y=self.find(x),self.find(y)
if x != y:
self.father[y]=x # 合并
self.count +=1
return 0
result = 0
nums = [x for x in range(7)]
for x in range(1,8): # 每次用的晶体管个数
for k in itertools.combinations(nums, x): #从所有的晶体管中按个数要求不重复的拿。
l = len(k)#晶体管的个数
ufs = UnionFindSet(l)
#两两的逐个选取
for a in range(l):
for b in range(a+1,l):
#根据下图的数字判断两个晶体管是否相邻。
if abs(k[a]-k[b])==1 or (k[a]==1 and k[b]==6) or (k[a]==2 and k[b]==6) or (k[a]==4 and k[b]==6) or (k[a]==0 and k[b]==5):
ufs.union(a,b)
if l-ufs.count==1: #比如当用到三个二极管的时候只需要链接两次,那么当晶体管个数减去链接次数为1的时候符合要求。
result += 1
print(result)
合根植物
合根植物:https://www.lanqiao.cn/problems/110/learning/
# https://www.lanqiao.cn/problems/110/learning/
m,n = map(int,input().split())
k = int(input())
f = [i for i in range(m*n+1)]
def find(x):
if f[x] != x:
f[x] = find(f[x]) # 路径压缩
return f[x]
def union(x,y):
fx = find(x)
fy = find(y)
if fx != fy:
f[fx] = f[fy]
for i in range(k):
x,y = map(int,input().split())
union(x, y)
ans = 0
for i in range(m*n):
if f[i] == i:
ans += 1
print(ans)
小猪存钱罐
小猪存钱罐 https://www.lanqiao.cn/problems/1085/learning/
# https://www.lanqiao.cn/problems/1085/learning/
def find(x):
if f[x] != x:
f[x] = find(f[x]) # 路径压缩
return f[x]
# 合并两个
def union(x,y):
fx = find(x)
fy = find(y)
if fx != fy:
f[fx] = f[fy]
N = int(input())
f = [i for i in range(N+1)]
for i in range(1,N+1):
union(i,int(input()))
ans = 0
for i in range(1,N+1):
if f[i] == i:
ans += 1
print(ans)
其他题目
火星旅行:https://www.lanqiao.cn/problems/1084/learning/
方格染色:https://www.lanqiao.cn/problems/1012/learning/
星球大战:https://www.lanqiao.cn/problems/828/learning/