Problem:


Score : 600 points

Problem Statement

There is a grid with N rows and N columns of squares. Let (i, j) be the square at the i-th row from the top and the j-th column from the left.

Each of the central (N-2) \times (N-2) squares in the grid has a black stone on it.
Each of the 2N - 1 squares on the bottom side and the right side has a white stone on it.

Q queries are given. We ask you to process them in order.
There are two kinds of queries. Their input format and description are as follows:

1 x: Place a white stone on (1, x). After that, for each black stone between (1, x) and the first white stone you hit if you go down from (1, x), replace it with a white stone.

2 x: Place a white stone on (x, 1). After that, for each black stone between (x, 1) and the first white stone you hit if you go right from (x, 1), replace it with a white stone.

How many black stones are there on the grid after processing all Q queries?

Constraints

3 \leq N \leq 2\times 10^5

0 \leq Q \leq \min(2N-4,2\times 10^5)

2 \leq x \leq N-1

Queries are pairwise distinct.

Input

Input is given from Standard Input in the following format:

N Q
Query_1
\vdots
Query_Q

Output

Print how many black stones there are on the grid after processing all Q queries.

Sample Input 1

5 5
1 3
2 3
1 4
2 2
1 2

Sample Output 1

1

After each query, the grid changes in the following way:

Sample Input 2

200000 0

Sample Output 2

39999200004

Sample Input 3

176527 15
1 81279
2 22308
2 133061
1 80744
2 44603
1 170938
2 139754
2 15220
1 172794
1 159290
2 156968
1 56426
2 77429
1 97459
2 71282

Sample Output 3

31159505795



Sample input:
5 5
1 3
2 3
1 4
2 2
1 2


Sample output:
1

Submission status: Accepted

Solution code:
class lazy_segtree:
  #遅延評価セグメント木
  def __init__(s, op, e, mapping, composition, id, v):
    if type(v) is int: v = [e()] * v
    s._n = len(v)
    s.log = s.ceil_pow2(s._n)
    s.size = 1 << s.log
    s.d = [e()] * (2 * s.size)
    s.lz = [id()] * s.size
    s.e = e
    s.op = op
    s.mapping = mapping
    s.composition = composition
    s.id = id
    for i in range(s._n): s.d[s.size + i] = v[i]
    for i in range(s.size - 1, 0, -1): s.update(i)
  
  # 1点更新
  def set(s, p, x):
    p += s.size
    for i in range(s.log, 0, -1): s.push(p >> i)
    s.d[p] = x
    for i in range(1, s.log + 1): s.update(p >> i)
 
  # 1点取得
  def get(s, p):
    p += s.size
    for i in range(s.log, 0, -1): s.push(p >> i)
    return s.d[p]
 
  # 区間演算
  def prod(s, l, r):
    if l == r: return s.e()
    l += s.size
    r += s.size
    for i in range(s.log, 0, -1):
      if (((l >> i) << i) != l): s.push(l >> i)
      if (((r >> i) << i) != r): s.push(r >> i)
    sml, smr = s.e(), s.e()
    while (l < r):
      if l & 1: 
        sml = s.op(sml, s.d[l])
        l += 1
      if r & 1:
        r -= 1
        smr = s.op(s.d[r], smr)
      l >>= 1
      r >>= 1
    return s.op(sml, smr)
 
  # 全体演算
  def all_prod(s): return s.d[1]
 
  # 1点写像
  def apply(s, p, f):
    p += s.size
    for i in range(s.log, 0, -1): s.push(p >> i)
    s.d[p] = s.mapping(f, s.d[p])
    for i in range(1, s.log + 1): s.update(p >> i)
 
  # 区間写像
  def apply(s, l, r, f):
    if l == r: return
    l += s.size
    r += s.size
    for i in range(s.log, 0, -1):
      if (((l >> i) << i) != l): s.push(l >> i)
      if (((r >> i) << i) != r): s.push((r - 1) >> i)
    l2, r2 = l, r
    while l < r:
      if l & 1: 
        sml = s.all_apply(l, f)
        l += 1
      if r & 1:
        r -= 1
        smr = s.all_apply(r, f)
      l >>= 1
      r >>= 1
    l, r = l2, r2
    for i in range(1, s.log + 1):
      if (((l >> i) << i) != l): s.update(l >> i)
      if (((r >> i) << i) != r): s.update((r - 1) >> i)
 
  def update(s, k): s.d[k] = s.op(s.d[2 * k], s.d[2 * k + 1])
  def all_apply(s, k, f):
    s.d[k] = s.mapping(f, s.d[k])
    if k < s.size: s.lz[k] = s.composition(f, s.lz[k])
  def push(s, k):
    s.all_apply(2 * k, s.lz[k])
    s.all_apply(2 * k + 1, s.lz[k])
    s.lz[k] = s.id()
  def ceil_pow2(s, n):
    x = 0
    while (1 << x) < n: x += 1
    return x

import sys
    
def e():
  return N - 1

def op(s, t):
  return min(s, t)

def mapping(f, a):
  return min(f, a)

def composition(f, g):
  return min(f, g)

def id():
  return N - 1

N, Q = list(map(int, input().split()))
Query = [list(map(int, input().split())) for _ in range(Q)]

v = [N - 2] * ((N - 2) * 2)


ans = (N - 2) * (N - 2)
seg = lazy_segtree(op, e, mapping, composition, id, v)

for q, x in Query:
  x -= 2
  l = 0
  r = 0
  t = 0
  if q == 1:
    t = N - 2
  if q == 2:
    l += N - 2
    r += N - 2
  g = seg.get(x + t)
  ans -= g
  r += g
  seg.apply(l, r, x)

print(ans)

Submission status: Accepted

Solution code:
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines

inf = 10**10

class Seg_min():
    def __init__(self,x):
        #####単位元######
        self.ide_ele_min = inf
        self.func = min

        self.n = len(x)

        #num_max:n以上の最小の2のべき乗
        self.num_max =2**(self.n-1).bit_length()
        self.x = [self.ide_ele_min]*2*self.num_max

        for i,num in enumerate(x, self.num_max):
            self.x[i] = num
        for i in range(self.num_max-1,0,-1):
            self.x[i] = self.func(self.x[i<<1],self.x[(i<<1) + 1])

    def update(self,i,x):
        i += self.num_max
        self.x[i] = x
        while(i>0):
            i = i//2
            self.x[i] = self.func(self.x[i<<1],self.x[(i<<1) + 1])

    def query(self,i,j):
        res = self.ide_ele_min
        if i>=j:
            return res
        i += self.num_max
        j += self.num_max -1
        while(i<=j):
            if(i==j):
                res = self.func(res,self.x[i])
                break
            if(i&1):
                res = self.func(res,self.x[i])
                i += 1
            if(not j&1):
                res = self.func(res,self.x[j])
                j -= 1
            i = i>>1
            j = j>>1
        return res

n,q,*query = list(map(int,read().split()))

wall_ns = Seg_min([inf] * (n-2))
wall_ew = Seg_min([inf] * (n-2))
min_ns = [n-2] * (q+1)
min_ew = [n-2] * (q+1)

ans = (n-2)*(n-2)
for i in range(1,q+1):
    x = query[i*2-1] - 2
    if(query[i*2-2] == 1):
        last = wall_ns.query(0,x)
        if(last == inf):
            last = i-1
        ans -= min_ew[last]

        wall_ns.update(x,i)
        min_ns[i] = min(min_ns[i-1], x)
        min_ew[i] = min_ew[i-1]
    else:
        last = wall_ew.query(0,x)
        if(last == inf):
            last = i-1
        ans -= min_ns[last]

        wall_ew.update(x,i)
        min_ew[i] = min(min_ew[i-1], x)
        min_ns[i] = min_ns[i-1]

print(ans)

Submission status: Accepted

Solution code:
import sys
input = lambda : sys.stdin.readline().rstrip()
sys.setrecursionlimit(max(1000, 10**9))
write = lambda x: sys.stdout.write(x+"\n")


### 遅延評価セグメント木
class LazySegmentTree:
    def __init__(self, n, a=None):
        """初期化
        num : n以上の最小の2のべき乗
        """
        num = 1
        while num<=n:
            num *= 2
        self.num = num
        self.seg = [ninf] * (2*self.num-1)
        self.lazy = [f0] * (2*self.num-1)
        self.ls = [0]*(2*self.num-1)
        self.rs = [0]*(2*self.num-1)
        self.ls[0] = 0
        self.rs[0] = self.num
        for i in range(self.num-1):
            self.ls[2*i+1] = self.ls[i]
            self.rs[2*i+1] = (self.ls[i] + self.rs[i])//2
            self.ls[2*i+2] = (self.ls[i] + self.rs[i])//2
            self.rs[2*i+2] = self.rs[i]
        if a is not None:
            # O(n)で初期化
            assert len(a)==n
            for i in range(n):
                self.seg[num-1+i] = a[i]
            for k in range(num-2, -1, -1):
                self.seg[k] = op(self.seg[2*k+1], self.seg[2*k+2])
    def eval(self, k):
        if self.lazy[k]==f0:
            return 
        if k<self.num-1:
            self.lazy[k*2+1] = composition(self.lazy[k], self.lazy[k*2+1])
            self.lazy[k*2+2] = composition(self.lazy[k], self.lazy[k*2+2])
        self.seg[k] = mapping(self.lazy[k], self.seg[k])
        self.lazy[k] = f0
    def eval_all(self):
        for i in range(2*self.num-1):
            self.eval(i)
    def update(self,a,b,x=None,f=None):
        """A[a]...A[b-1]をxに更新する
        """
        if f is None:
            # 更新クエリ
            f = lambda y: x
        k = 0
        q = [k] # k>=0なら行きがけ順
        # 重なる区間を深さ優先探索
        while q:
            k = q.pop()
            l,r = self.ls[k], self.rs[k]
            if k>=0:
                self.eval(k)
                if r<=a or b<=l:
                    continue
                elif a<=l and r<=b:
                    self.lazy[k] = composition(f, self.lazy[k])
                    self.eval(k)
                else:
                    q.append(~k)
                    q.append(2*k+1)
                    q.append(2*k+2)
            else:
                k = ~k
                self.seg[k] = op(self.seg[2*k+1], self.seg[2*k+2])
    def query(self,a,b):
        k = 0
        l = 0
        r = self.num
        q = [k]
        ans = ninf
        # 重なる区間を深さ優先探索
        while q:
            k = q.pop()
            l,r = self.ls[k], self.rs[k]
            self.eval(k)
            if r<=a or b<=l:
                continue
            elif a<=l and r<=b:
                ans = op(ans, self.seg[k])
            else:
                q.append(2*k+2)
                q.append(2*k+1)
#             print(q, ans, l,r,a,b, self.seg[k])
        return ans

n,q = list(map(int, input().split()))

ninf = 10**15
op = lambda x,y: min(x,y)
mapping = lambda f,x: min(f,x) if f is not None else x
def composition(f1,f2):
    if f1 is not None:
        if f2 is not None:
            return min(f1,f2)
        else:
            return f1
    else:
        return f2
# composition = lambda f1, f2: min(f1,f2) if f2 is not None else f1
f0 = None
sgr = LazySegmentTree(n-2, [n]*(n-2))
sgc = LazySegmentTree(n-2, [n]*(n-2))
ans = 0
for _ in range(q):
    t,x = map(int, input().split())
    x -= 2
    if t==1:
        v = sgc.query(x,x+1)
        ans += (v-2)
        sgr.update(0,v-2,f=x+2)
    else:
        v = sgr.query(x,x+1)
        ans += (v-2)
        sgc.update(0,v-2,f=x+2)
#     print(v-2)
ans = (n-2)**2 - ans
print(ans)

Submission status: Accepted

Solution code:
N,Q = map(int, input().split())
answer = (N-2)**2

q1 = [(N,N-2)]
q2 = [(N,N-2)]

def binary_search(q,n):
  left = 0
  right = len(q) - 1
  while right > left:
    index = (left + right) // 2
    if q[index+1][0] > n:
      left = index + 1
    else:
      right = index
  return left

for _ in range(Q):
  n,x = map(int, input().split())
  if n == 1:
    index = binary_search(q1,x)
    q = q1[index]
    answer -= q[1]
    if index == len(q1)-1:
      q1.append((x,q[1]))
      q2_last = q2[-1]
      q2[-1] = (q2_last[0],x-2)
  if n == 2:
    index = binary_search(q2,x)
    q = q2[index]
    answer -= q[1]
    if index == len(q2)-1:
      q2.append((x,q[1]))
      q1_last = q1[-1]
      q1[-1] = (q1_last[0],x-2)
      
print(answer)

Submission status: Accepted

Solution code:
from bisect import bisect

N, Q = map(int, input().split())

ans = (N-2)*(N-2)
wx = wy = 0
Ax = []
Ay = []

for _ in range(Q):
    t, p = map(int, input().split())
    p = N-p
    
    if t==1:
        i = bisect(Ax, (p, 0))
        if i==len(Ax):
            Ax.append((p, wy))
            wx = p
            rev = N-2-wy
        else:
            rev = N-2-Ax[i][1]
    else:
        i = bisect(Ay, (p, 0))
        if i==len(Ay):
            Ay.append((p, wx))
            wy = p
            rev = N-2-wx
        else:
            rev = N-2-Ay[i][1]

    ans -= rev

print(ans)