[
segment tree
]
CodeForces Segment Trees, Part 2, 2.5
Problem statement
https://codeforces.com/edu/course/2/lesson/5/2/practice/contest/279653/problem/E
Solution
This segment tree will support two operations:
- Apply set with value
v
for all elements on[l, r)
- Find
min
for all elements on[l, r)
Difference from previous trees is that we can use not-commutative operation for this we will use propagation.
Complexity
Standart compexities for segment tree.
Code
class SegmentTree:
def __init__(self, n):
self.size = 1
while self.size < n:
self.size *= 2
self.T = [0] * (2 * self.size - 1) # to multiply
self.L = [0] * (2 * self.size - 1) # current sum
self.ZERO = float("inf") # neutral element, for min it is +inf
self.NO_OPERATION = -float("inf") # this is for our propagate to understand that we do not need it
def op_modify(self, a, b, len_):
if b == self.NO_OPERATION:
return a
return b
def op_sum(self, a, b):
return min(a, b)
def propagate(self, x, lx, rx):
if self.L[x] == self.NO_OPERATION or rx - lx == 1:
return
mx = (lx + rx)//2
self.L[2 * x + 1] = self.op_modify(self.L[2 * x + 1], self.L[x], 1)
self.T[2 * x + 1] = self.op_modify(self.T[2 * x + 1], self.L[x], mx - lx) # in fact we do not use len
self.L[2 * x + 2] = self.op_modify(self.L[2 * x + 2], self.L[x], 1)
self.T[2 * x + 2] = self.op_modify(self.T[2 * x + 2], self.L[x], rx - mx)
self.L[x] = self.NO_OPERATION
def _update(self, l, r, v, x, lx, rx):
self.propagate(x, lx, rx)
if l >= rx or lx >= r:
return
if lx >= l and rx <= r:
self.L[x] = self.op_modify(self.L[x], v, 1) # why 1?
self.T[x] = self.op_modify(self.T[x], v, rx - lx)
return
mx = (lx + rx)//2
self._update(l, r, v, 2*x+1, lx, mx)
self._update(l, r, v, 2*x+2, mx, rx)
self.T[x] = self.op_sum(self.T[2*x+1], self.T[2*x+2])
def update(self, l, r, v):
return self._update(l, r, v, 0, 0, self.size)
def _query(self, l, r, x, lx, rx):
self.propagate(x, lx, rx)
if l >= rx or lx >= r:
return self.ZERO
if lx >= l and rx <= r:
return self.T[x]
mx = (lx + rx) // 2
m1 = self._query(l, r, 2 * x + 1, lx, mx)
m2 = self._query(l, r, 2 * x + 2, mx, rx)
return self.op_sum(m1, m2)
def query(self, l, r):
return self._query(l, r, 0, 0, self.size)
if __name__ == '__main__':
n, m = [int(i) for i in input().split()]
MOD = 10**9 + 7
STree = SegmentTree(n)
for i in range(m):
t = [int(i) for i in input().split()]
if t[0] == 1:
STree.update(t[1], t[2], t[3])
else:
print(STree.query(t[1], t[2]))