[
segment tree
]
CodeForces Segment Trees, Part 2, 2.1
Problem statement
https://codeforces.com/edu/course/2/lesson/5/2/practice/contest/279653/problem/A
Solution
This segment tree will support two operations:
- add value
v
to all elements in range[l, r)
- find
min
for all elements in range[l, r)
Now in tree we keep two values: what we need to add
and current min
. Here we do not use propagation, because operations are commutative.
Complexity
Standart compexities for segment tree.
Code
class SegmentTree:
def __init__(self, n, arr):
self.size = 1
while self.size < n:
self.size *= 2
self.T = [0] * (2 * self.size - 1) # to add
self.L = [0] * (2 * self.size - 1) # current min
self.arr = arr
def _add(self, l, r, v, x, lx, rx):
if l >= rx or lx >= r:
return
if lx >= l and rx <= r:
self.T[x] += v
self.L[x] += v
return
mx = (lx + rx)//2
self._add(l, r, v, 2*x+1, lx, mx)
self._add(l, r, v, 2*x+2, mx, rx)
self.T[x] = min(self.T[2*x+1], self.T[2*x+2]) + self.L[x]
def add(self, l, r, v):
return self._add(l, r, v, 0, 0, self.size)
def _min(self, l, r, x, lx, rx):
if l >= rx or lx >= r:
return float("inf")
if lx >= l and rx <= r:
return self.T[x]
mx = (lx + rx) // 2
m1 = self._min(l, r, 2 * x + 1, lx, mx)
m2 = self._min(l, r, 2 * x + 2, mx, rx)
return min(m1, m2) + self.L[x]
def min(self, l, r):
return self._min(l, r, 0, 0, self.size)
if __name__ == '__main__':
n, m = [int(i) for i in input().split()]
STree = SegmentTree(n, [0]*n)
for i in range(m):
t = [int(i) for i in input().split()]
if t[0] == 1:
STree.add(t[1], t[2], t[3])
else:
print(STree.min(t[1], t[2]))