前言
整体评价
今天相对容易，E的构造题，感谢出题人极其善意的Case 1, 算是放水了。F题是个很典的结论题，由于存在动态点修改，所以引入树状数组做区间和的快速计算。
A. 小红的数位删除
题型: 签到
s = input()
print (s[:-3])
B. 小红的小红矩阵构造
思路: 模拟
h, w, s = list(map(int, input().split()))
ts = 0
grid = []
for i in range(h):
arr = list(map(int, input().split()))
grid.append(arr)
ts += sum(arr)
hs = set()
for i in range(h):
v = 0
for u in grid[i]:
v = v ^ u
hs.add(v)
for i in range(w):
v = 0
for j in range(h):
v ^= grid[j][i]
hs.add(v)
if len(hs) == 1 and ts == s:
print ("accepted")
else:
print ("wrong answer")
C. 小红的白色字符串
思路: 状态机DP
引入三个状态
- 0, 末尾是空格，
- 1, 末尾是小写字母
- 2, 末尾是大写字母
具体转移和当前的字母的大小写性质有关
具体看代码
s = input()
from math import inf
dp0, dp1, dp2 = 0, inf, inf
n = len(s)
for c in s:
if c >= 'A' and c <= 'Z':
t0 = min(dp0, dp1, dp2) + 1
t1 = inf
t2 = dp0
dp0, dp1, dp2 = t0, t1, t2
else:
t0 = min(dp0, dp1, dp2) + 1
t1 = min(dp0, dp1, dp2)
t2 = inf
dp0, dp1, dp2 = t0, t1, t2
print (min(dp0, dp1, dp2))
D. 小红走矩阵
思路: BFS
很板的一道题，只是在扩展下一步的时候，多了一个限制条件
h, w = list(map(int, input().split()))
maze = []
for _ in range(h):
arr = list(input())
maze.append(arr)
from math import inf
from collections import deque
vis = [[inf] * w for _ in range(h)]
deq = deque()
deq.append((0, 0))
vis[0][0] = 0
while len(deq) > 0:
y, x = deq.popleft()
for (dy, dx) in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
ty = y + dy
tx = x + dx
if 0 <= ty < h and 0 <= tx < w:
if maze[y][x] != maze[ty][tx]:
if vis[ty][tx] > vis[y][x] + 1:
vis[ty][tx] = vis[y][x] + 1
deq.append((ty, tx))
if vis[h - 1][w - 1] == inf:
print (-1)
else:
print (vis[h - 1][w - 1])
E. 小红的小红走矩阵
思路: 构造一个弯
按照Case 1，构造一个”钩“，然后其余的保持和周边不同即可。
因为地图四色定律，所以只要”abcd“即可。
h, w = list(map(int, input().split()))
maze = [['A'] * w for _ in range(h)]
maze[0][0] = 'a'
maze[0][1] = 'b'
maze[0][2] = 'b'
maze[1][0] = 'a'
maze[1][1] = 'c'
maze[1][2] = 'c'
maze[2][0] = 'b'
maze[2][1] = 'c'
def compute(y, x):
s = set()
for (dy, dx) in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
ty, tx = y + dy, x + dx
if 0 <= ty < h and 0 <= tx < w:
if maze[ty][tx] != 'A':
s.add(maze[ty][tx])
for c in "abcd":
if c not in s:
return c
return 'e'
for i in range(h):
for j in range(w):
if maze[i][j] == 'A':
maze[i][j] = compute(i, j)
for i in range(h):
print (''.join(maze[i]))
F. 小红的好子串询问
思路: 经典结论 + 树状数组
不存在 大于等于2个子串的回文，
r e d 的排列 p ，然后 p 循环出现 red的排列p，然后p循环出现 red的排列p，然后p循环出现
基于这个结论，对于某个查询[l, r], 则枚举red的排列，然后引入9个树状数组, 进行统计，求最小代价即可。
fenwicks[3][3]，表示位置余3，字母(按red=123）的9个树状数组
class BIT(object):
def __init__(self, n):
self.n = n
self.arr = [0] * (n + 1)
def query(self, p):
r = 0
while p > 0:
r += self.arr[p]
p -= p & -p
return r
def update(self, p, d):
while p <= self.n:
self.arr[p] += d
p += p & -p
def toId(ch):
if ch == 'r':
return 0
elif ch == 'e':
return 1
return 2
from math import inf
from itertools import permutations
n, m = list(map(int, input().split()))
s = list(input())
trees = [ [ BIT(n) for _ in range(3) ] for _ in range(3)]
for i in range(n):
trees[i % 3][toId(s[i])].update(i + 1, 1)
for _ in range(m):
op, a1, a2 = input().split()
if int(op) == 1:
l, ch = int(a1) - 1, a2[0]
trees[l % 3][toId(s[l])].update(l + 1, -1)
s[l] = ch
trees[l % 3][toId(s[l])].update(l + 1, 1)
else:
l, r = int(a1) - 1, int(a2) - 1
res = inf
for (t1, t2, t3) in list(permutations([0, 1, 2])):
z1 = trees[(l + t1) % 3][0].query(r + 1) - trees[(l + t1) % 3][0].query(l)
z2 = trees[(l + t2) % 3][1].query(r + 1) - trees[(l + t2) % 3][1].query(l)
z3 = trees[(l + t3) % 3][2].query(r + 1) - trees[(l + t3) % 3][2].query(l)
res = min(res, (r - l + 1) - (z1 + z2 + z3))
print (res)
写在最后
