AtCoder abc130
F题提交了无数遍，最后发现是三分求解的写法错了
C - Rectangle Cutting
盲猜都在xy的中心点时可以无限分割，否则不能
D - Enough Array
前缀和二分求位置
E - Common Subsequence
公共子序列求有几种组合
设 d p [ i ] [ j ] dp[i][j] dp[i][j]代表s取到i t取到j时的序列数
当s[i]!=t[j] 时
d p [ i ] [ j ] = d p [ i − 1 ] [ j ] + d p [ i ] [ j − 1 ] − d p [ i − 1 ] [ j − 1 ] dp[i][j]=dp[i-1] [j] + dp[i][j - 1] - dp[i - 1][j - 1] dp[i][j]=dp[i−1][j]+dp[i][j−1]−dp[i−1][j−1]
因为 d p [ i ] [ j ] dp[i][j] dp[i][j]可以视作为 d p [ i − 1 ] [ j ] dp[i - 1][j] dp[i−1][j]添上s[i]后总的序列数
d p [ i − 1 ] [ j ] dp[i-1][j] dp[i−1][j]是 d p [ i − 1 ] [ j − 1 ] dp[i-1][j-1] dp[i−1][j−1]添上t[j]的序列数
另一边 d p [ i ] [ j − 1 ] dp[i][j - 1] dp[i][j−1]也将 d p [ i − 1 ] [ j − 1 ] dp[i-1][j-1] dp[i−1][j−1]包含在内，因此计算了两次 d p [ i − 1 ] [ j − 1 ] dp[i-1][j-1] dp[i−1][j−1]需要减去
当s[i]==t[j]时， d p [ i ] [ j ] dp[i][j] dp[i][j]在 d p [ i − 1 ] [ j − 1 ] dp[i-1][j-1] dp[i−1][j−1]的序列上各增加一个长度，因此在刚才的计算后再加上 d p [ i − 1 ] [ j − 1 ] dp[i-1][j-1] dp[i−1][j−1]
# -*- coding: utf-8 -*-
# @time : 2023/6/2 13:30
# @author : [email protected]
# @desc :
# @file : atcoder.py
# @software : PyCharm
import bisect
import copy
import sys
from sortedcontainers import SortedList
from collections import defaultdict, Counter, deque
from functools import lru_cache, cmp_to_key
import heapq
import math
sys.setrecursionlimit(100010)
mod = 10 ** 9 + 7
def main():
items = sys.version.split()
if items[0] == '3.10.6':
fp = open("in.txt")
else:
fp = sys.stdin
n, m = map(int, fp.readline().split())
s = list(map(int, fp.readline().split()))
t = list(map(int, fp.readline().split()))
dp = [[0] * (m + 1) for _ in range(n + 1)]
for i in range(n + 1):
dp[i][0] = 1
for i in range(m + 1):
dp[0][i] = 1
for i in range(1, n + 1):
for j in range(1, m + 1):
if s[i - 1] == t[j - 1]:
dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
else:
dp[i][j] = dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1]
dp[i][j] %= mod
print(dp[n][m])
if __name__ == "__main__":
main()
F - Minimum Bounding Box
max-min显然是凸函数（忘了证明方法），暴力三分可以过
还有一种不那么暴力的解法：
不需要维护所有的x y
只需要维护向上、向下的y中最大值与最小值
向左向右x最大值与最小值
# -*- coding: utf-8 -*-
# @time : 2023/6/2 13:30
# @author : [email protected]
# @desc :
# @file : atcoder.py
# @software : PyCharm
import bisect
import copy
import sys
from sortedcontainers import SortedList
from collections import defaultdict, Counter, deque
from functools import lru_cache, cmp_to_key
import heapq
import math
sys.setrecursionlimit(100010)
def main():
items = sys.version.split()
if items[0] == '3.10.6':
fp = open("in.txt")
else:
fp = sys.stdin
n = int(fp.readline())
min_x, min_y, max_x, max_y = 10 ** 20, 10 ** 20, -10 ** 20, -10 ** 20
uy, dy, lx, rx = [], [], [], []
for i in range(n):
items = fp.readline().strip().split()
x, y = int(items[0]), int(items[1])
d = items[2]
if d == 'U':
uy.append(y)
min_x, max_x = min(min_x, x), max(max_x, x)
elif d == 'D':
dy.append(y)
min_x, max_x = min(min_x, x), max(max_x, x)
elif d == 'L':
lx.append(x)
min_y, max_y = min(min_y, y), max(max_y, y)
else:
rx.append(x)
min_y, max_y = min(min_y, y), max(max_y, y)
uy.sort()
dy.sort()
lx.sort()
rx.sort()
def get(t):
x0, y0, x1, y1 = min_x, min_y, max_x, max_y
if len(uy) > 0:
y0 = min(uy[0] + t, y0)
y1 = max(uy[-1] + t, y1)
if len(dy) > 0:
y0 = min(dy[0] - t, y0)
y1 = max(dy[-1] - t, y1)
if len(rx) > 0:
x0 = min(rx[0] + t, x0)
x1 = max(rx[-1] + t, x1)
if len(lx) > 0:
x0 = min(lx[0] - t, x0)
x1 = max(lx[-1] - t, x1)
return (y1 - y0) * (x1 - x0)
lo, hi = 0, 10 ** 13
c = 0
ans = 1e18
while c < 400:
m0, m1 = lo + (hi - lo) / 3, hi - (hi - lo) / 3
a0, a1 = get(m0), get(m1)
if a0 > a1:
lo = m0
else:
hi = m1
ans = min(ans, a0)
ans = min(ans, a1)
c += 1
print(ans)
if __name__ == "__main__":
main()