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386 lines
10 KiB
Python
Executable file
386 lines
10 KiB
Python
Executable file
"""#
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### 谜题描述
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There are n points on the plane, the i-th of which is at (x_i, y_i). Tokitsukaze wants to draw a strange rectangular area and pick all the points in the area.
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The strange area is enclosed by three lines, x = l, y = a and x = r, as its left side, its bottom side and its right side respectively, where l, r and a can be any real numbers satisfying that l < r. The upper side of the area is boundless, which you can regard as a line parallel to the x-axis at infinity. The following figure shows a strange rectangular area.
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<image>
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A point (x_i, y_i) is in the strange rectangular area if and only if l < x_i < r and y_i > a. For example, in the above figure, p_1 is in the area while p_2 is not.
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Tokitsukaze wants to know how many different non-empty sets she can obtain by picking all the points in a strange rectangular area, where we think two sets are different if there exists at least one point in one set of them but not in the other.
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Input
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The first line contains a single integer n (1 ≤ n ≤ 2 × 10^5) — the number of points on the plane.
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The i-th of the next n lines contains two integers x_i, y_i (1 ≤ x_i, y_i ≤ 10^9) — the coordinates of the i-th point.
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All points are distinct.
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Output
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Print a single integer — the number of different non-empty sets of points she can obtain.
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Examples
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Input
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3
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1 1
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1 2
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1 3
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Output
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3
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Input
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3
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1 1
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2 1
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3 1
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Output
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6
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Input
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4
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2 1
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2 2
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3 1
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3 2
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Output
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6
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Note
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For the first example, there is exactly one set having k points for k = 1, 2, 3, so the total number is 3.
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For the second example, the numbers of sets having k points for k = 1, 2, 3 are 3, 2, 1 respectively, and their sum is 6.
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For the third example, as the following figure shows, there are
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* 2 sets having one point;
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* 3 sets having two points;
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* 1 set having four points.
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Therefore, the number of different non-empty sets in this example is 2 + 3 + 0 + 1 = 6.
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<image>
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Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
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```python
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class seg:
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def __init__(self,n):
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self.m = 1
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while self.m < n: self.m *= 2
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self.data = [0]*(2*self.m)
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def add(self, ind, val):
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ind += self.m
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while ind:
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self.data[ind] += val
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ind >>= 1
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def summa(self, l, r):
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l += self.m
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r += self.m
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s = 0
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while l<r:
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if l&1:
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s += self.data[l]
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l += 1
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if r&1:
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r -= 1
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s += self.data[r]
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l >>= 1
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r >>= 1
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return s
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def main():
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inp = readnumbers()
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ii = 0
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n = inp[ii]
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ii += 1
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X = inp[ii :ii + 2*n:2]
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Y = inp[ii + 1:ii + 2*n:2]
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ii += 2*n
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X2 = sorted(X)
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Y2 = sorted(Y)
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xmap = {}
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for x in X2:
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if x not in xmap:
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xmap[x] = len(xmap)
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ymap = {}
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for y in Y2:
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if y not in ymap:
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ymap[y] = len(ymap)
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X = [xmap[x] for x in X]
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Y = [ymap[y] for y in Y]
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Ybuckets = [[] for _ in range(len(ymap))]
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for i in range(n):
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Ybuckets[Y[i]].append(X[i])
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for yvec in Ybuckets:
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yvec.sort()
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ways = 0
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sumseg = seg(len(xmap))
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for yvec in reversed(Ybuckets):
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for x in yvec:
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if sumseg.summa(x, x+1)==0:
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sumseg.add(x,1)
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old_x = -1
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for x in yvec:
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ways += sumseg.summa(old_x + 1, x + 1) * (sumseg.summa(x + 1, len(xmap)) + 1)
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old_x = x
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print ways
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######## Python 2 and 3 footer by Pajenegod and c1729
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# Note because cf runs old PyPy3 version which doesn't have the sped up
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# unicode strings, PyPy3 strings will many times be slower than pypy2.
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# There is a way to get around this by using binary strings in PyPy3
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# but its syntax is different which makes it kind of a mess to use.
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# So on cf, use PyPy2 for best string performance.
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py2 = round(0.5)
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if py2:
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from future_builtins import ascii, filter, hex, map, oct, zip
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range = xrange
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import os, sys
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from io import IOBase, BytesIO
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BUFSIZE = 8192
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class FastIO(BytesIO):
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newlines = 0
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def __init__(self, file):
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self._file = file
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self._fd = file.fileno()
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self.writable = \"x\" in file.mode or \"w\" in file.mode
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self.write = super(FastIO, self).write if self.writable else None
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def _fill(self):
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s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
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self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
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return s
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def read(self):
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while self._fill(): pass
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return super(FastIO,self).read()
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def readline(self):
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while self.newlines == 0:
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s = self._fill(); self.newlines = s.count(b\"\n\") + (not s)
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self.newlines -= 1
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return super(FastIO, self).readline()
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def flush(self):
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if self.writable:
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os.write(self._fd, self.getvalue())
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self.truncate(0), self.seek(0)
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class IOWrapper(IOBase):
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def __init__(self, file):
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self.buffer = FastIO(file)
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self.flush = self.buffer.flush
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self.writable = self.buffer.writable
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if py2:
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self.write = self.buffer.write
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self.read = self.buffer.read
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self.readline = self.buffer.readline
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else:
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self.write = lambda s:self.buffer.write(s.encode('ascii'))
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self.read = lambda:self.buffer.read().decode('ascii')
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self.readline = lambda:self.buffer.readline().decode('ascii')
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sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
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input = lambda: sys.stdin.readline().rstrip('\r\n')
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# Cout implemented in Python
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import sys
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class ostream:
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def __lshift__(self,a):
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sys.stdout.write(str(a))
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return self
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cout = ostream()
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endl = '\n'
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# Read all remaining integers in stdin, type is given by optional argument, this is fast
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def readnumbers(zero = 0):
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conv = ord if py2 else lambda x:x
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A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
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try:
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while True:
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if s[i] >= b'0' [0]:
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numb = 10 * numb + conv(s[i]) - 48
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elif s[i] == b'-' [0]: sign = -1
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elif s[i] != b'\r' [0]:
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A.append(sign*numb)
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numb = zero; sign = 1
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i += 1
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except:pass
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if s and s[-1] >= b'0' [0]:
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A.append(sign*numb)
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return A
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if __name__== \"__main__\":
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main()
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```
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请完成上述谜题的训练场环境类实现，包括所有必要的方法。
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"""
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from bootcamp import Basebootcamp
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import random
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from bootcamp import Basebootcamp
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class SegmentTree:
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def __init__(self, size):
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self.m = 1
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while self.m < size:
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self.m <<= 1
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self.data = [0] * (2 * self.m)
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def update(self, index, value):
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index += self.m
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while index > 0:
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self.data[index] += value
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index >>= 1
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def query_range(self, l, r):
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res = 0
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l += self.m
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r += self.m
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while l < r:
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if l % 2 == 1:
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res += self.data[l]
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l += 1
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if r % 2 == 1:
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r -= 1
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res += self.data[r]
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l >>= 1
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r >>= 1
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return res
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def calculate_answer(points):
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if not points:
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return 0
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# 离散化坐标
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x_coords = sorted({x for x, y in points})
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y_coords = sorted({y for x, y in points})
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x_map = {x: i for i, x in enumerate(x_coords)}
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y_map = {y: i for i, y in enumerate(y_coords)}
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# 按y分层存储x坐标
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y_buckets = [[] for _ in range(len(y_coords))]
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for x, y in points:
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y_idx = y_map[y]
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y_buckets[y_idx].append(x_map[x])
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for bucket in y_buckets:
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bucket.sort()
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total = 0
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st = SegmentTree(len(x_coords))
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# 按y降序处理
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for bucket in reversed(y_buckets):
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# 添加当前层的点
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for x in bucket:
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if st.query_range(x, x+1) == 0:
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st.update(x, 1)
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prev_x = -1
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for x in bucket:
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# 计算左区域贡献
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left = st.query_range(prev_x + 1, x + 1)
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# 计算右区域贡献（包括无穷大情况）
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right = st.query_range(x + 1, len(x_coords)) + 1
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total += left * right
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prev_x = x
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return total
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class Ftokitsukazeandstrangerectanglebootcamp(Basebootcamp):
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def __init__(self, min_n=1, max_n=10, x_max=100, y_max=100):
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self.min_n = min_n
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self.max_n = max_n
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self.x_max = x_max
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self.y_max = y_max
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def case_generator(self):
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n = random.randint(self.min_n, self.max_n)
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points = set()
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while len(points) < n:
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x = random.randint(1, self.x_max)
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y = random.randint(1, self.y_max)
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points.add((x, y))
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points = list(points)
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answer = calculate_answer(points)
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return {
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'n': n,
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'points': points,
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'answer': answer
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}
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@staticmethod
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def prompt_func(question_case):
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points = question_case['points']
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n = question_case['n']
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input_lines = [f"{x} {y}" for x, y in points]
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input_str = '\n'.join([str(n)] + input_lines)
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return f"""请解决以下几何谜题，输出最终答案的数值。
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题目描述：
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平面上有{n}个互不相同的点。定义一个特殊矩形区域：由三条直线x=l（左边界）、x=r（右边界，满足l<r）和y=a（底边界）围成，顶部无限延伸。点(x_i, y_i)位于区域内当且仅当l < x_i < r且y_i > a。求可以形成的不同非空点集的数量。
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输入格式：
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第一行：n
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接下来n行：每行两个整数x_i y_i
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输入数据：
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{input_str}
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请将答案数值置于[answer]标签内，例如：[answer]42[/answer]。确保结果为整数。"""
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@staticmethod
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def extract_output(output):
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import re
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matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
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if not matches:
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return None
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try:
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return int(matches[-1].strip())
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except (ValueError, TypeError):
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return None
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@classmethod
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def _verify_correction(cls, solution, identity):
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return solution == identity['answer']
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