Yukicoder No.325 マンハッタン距離2

,

http://yukicoder.me/problems/no/325

実装がとてもつらかった。やりかたが悪いのだろうか。

solution

やる。包除原理あたりを使って領域ごとに足し引きをする。原点や軸上に注意。

implementation

  • 原点
  • 軸上
  • それ以外

を求める関数をそれぞれ作って足し合わせた。

#!/usr/bin/env python3

# functions
def f(x, y, d): # (0, x) * (0, y)
    assert 0 <= x
    assert 0 <= y
    if x == 0 or y == 0:
        return 0
    x -= 1
    y -= 1
    d = min(d, x + y)
    ans = (d+1)*(d+2)//2
    if x <= d:
        dx = d-x
        ans -= dx*(dx+1)//2
    if y <= d:
        dy = d-y
        ans -= dy*(dy+1)//2
    ans -= min(x,d)+1
    ans -= min(y,d)+1
    ans += 1
    return ans
def g(x1, x2, y1, y2, d): # on the axes
    if x1 <= 0 <= x2:
        if 0 <= y1 <= y2:
            if y1 <= d:
                ans = min(d, y2) - min(d, y1) + 1
                if y1 == 0:
                    ans -= 1
                return ans
            else:
                return 0
        elif y1 <= 0 <= y2:
            return min(d, abs(y1)) + min(d, y2)
        assert False
    else:
        return 0
def h(x1, x2, y1, y2, d): # on the origin
    if x1 <= 0 <= x2 and y1 <= 0 <= y2:
        return 1
    else:
        return 0

# input
x1, y1, x2, y2, d = map(int,input().split())
assert x1 < x2
assert y1 < y2
if x2 <= 0:
    x1, x2 = - x2, - x1
if y2 <= 0:
    y1, y2 = - y2, - y1
if x1 <= 0:
    x1, y1 = y1, x1
    x2, y2 = y2, x2

# calc
ans = 0
if x1 <= 0:
    assert x1 <= 0 <= x2
    assert y1 <= 0 <= y2
    ans += f(abs(x1)+1, abs(y1)+1, d)
    ans += f(abs(x1)+1, abs(y2)+1, d)
    ans += f(abs(x2)+1, abs(y1)+1, d)
    ans += f(abs(x2)+1, abs(y2)+1, d)
elif y1 <= 0:
    assert 0 <= x1 < x2
    assert y1 <= 0 <= y2
    ans -= f(x1,   abs(y1)+1, d)
    ans += f(x2+1, abs(y1)+1, d)
    ans -= f(x1,   abs(y2)+1, d)
    ans += f(x2+1, abs(y2)+1, d)
else:
    assert 0 <= x1 < x2
    assert 0 <= y1 < y2
    ans += f(x1,   y1,   d)
    ans -= f(x1,   y2+1, d)
    ans -= f(x2+1, y1,   d)
    ans += f(x2+1, y2+1, d)
ans += g(x1, x2, y1, y2, d)
ans += g(y1, y2, x1, x2, d)
ans += h(x1, x2, y1, y2, d)

# output
print(ans)