标签：spl for 大连 rate == split() 最大 转移 __name__
以HDU1231为例，代码之没法交如下：
inf = 0x3f3f3f3f
a = [0 for i in range(10005)]
ans, L, R = -inf, 0, 0
def divide_and_conquer(l, r):
global ans, L, R
if l > r:
return
if l == r:
if ans < a[l]:
ans = a[l]
L, R = l, r
elif ans == a[l] and l < L:
L, R = l, r
return
mid = (l + r) // 2
divide_and_conquer(l, mid-1)
divide_and_conquer(mid+1, r)
tmp = a[mid]
l_ans, r_ans = 0, 0 # 左、右边部分的最大连续和(不含a[mid])
lL, rR = mid, mid
ll, rr = mid-1, mid+1
l_tmp, r_tmp = a[ll], a[rr]
while ll >= l and tmp+l_tmp > 0:
if l_ans < l_tmp:
l_ans = l_tmp
lL = ll
ll -= 1
l_tmp += a[ll]
while rr <= r and tmp+r_tmp > 0:
if r_ans < r_tmp:
r_ans = r_tmp
rR = rr
rr += 1
r_tmp += a[rr]
if ans < tmp + l_ans + r_ans:
ans = tmp + l_ans + r_ans
L, R = lL, rR
elif ans == tmp + l_ans + r_ans and lL < L:
L, R = lL, rR
if __name__ == ‘__main__‘:
k = int(input())
for i, e in enumerate(list(map(lambda x: int(x), input().split()))):
a[i+1] = e
divide_and_conquer(1, k)
if ans < 0:
L, R = 1, k
print(ans, a[L], a[R])
DP方法就较为简单了，状态dp[i]表示以第i个元素作为结尾的连续子序列的最大和，转移：遍历时，若前一个状态小于0则dp[i] = e[i]，否则dp[i] = dp[i-1] + e[i]。
go on~
divide and conquer - 最大连续子序列 - py
标签：spl for 大连 rate == split() 最大 转移 __name__
原文地址：https://www.cnblogs.com/boceng/p/11502708.html
