题目 | One Fourth

AtCoder Beginner Contest 250

F - One Fourth

https://atcoder.jp/contests/abc250/tasks/abc250_f

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : $500$ points

Problem Statement

ABC 250 is a commemorable quarter milestone for Takahashi, who aims to hold ABC 1000, so he is going to celebrate this contest by eating as close to $1/4$ of a pizza he bought as possible.

The pizza that Takahashi bought has a planar shape of convex $N$-gon. When the pizza is placed on an $xy$-plane, the $i$-th vertex has coordinates $(X_i,Y_i)$.

Takahashi has decided to cut and eat the pizza as follows.

• First, Takahashi chooses two non-adjacent vertices from the vertices of the pizza and makes a cut with a knife along the line passing through those two points, dividing the pizza into two pieces.
• Then, he chooses one of the pieces at his choice and eats it.

Let aa be the quarter ($=1/4$) of the area of the pizza that Takahashi bought, and bb be the area of the piece of pizza that Takahashi eats. Find the minimum possible value of $8\times |a-b|$. We can prove that this value is always an integer.

Constraints

• All values in input are integers.
• $4\le N\le {10}^5$
• $|X_i|,|Y_i|\le 4\times {10}^8$
• The given points are the vertices of a convex $N$-gon in the counterclockwise order.

---

Input

Input is given from Standard Input in the following format:

$N$
$X_1$ $Y_1$
$X_2$ $Y_2$
$\dots$
$X_N$ $Y_N$

Output

Print the answer as an integer.

---

Sample Input 1

5
3 0
2 3
-1 3
-3 1
-1 -1

Sample Output 1

1

Suppose that he makes a cut along the line passing through the $3$-rd and the $5$-th vertex and eats the piece containing the $4$-th vertex.
Then, $a=\frac{33}{2}\times \frac{1}{4}=\frac{33}{8}$​, $b=4$, and $8\times |a-b|=1$, which is minimum possible.

---

Sample Input 2

4
400000000 400000000
-400000000 400000000
-400000000 -400000000
400000000 -400000000

Sample Output 2

1280000000000000000

---

Sample Input 3

6
-816 222
-801 -757
-165 -411
733 131
835 711
-374 979

Sample Output 3

157889

我的笔记

$O(N^3)$ 法：TLE x20

思路：

在 $P_0,P_1,\cdots ,P_N$ (P代表点) 中任取 $P_i,P_j$，且 $P_i,P_j$ 不可相邻，规定吃掉包含 $P_{i+1}$ 的部分。所有情况 $8\times |a-b|$ 的最小值即为所求。

求法：

$P_i,P_j$ 不相邻判断：$i,j$ 相差的绝对值不可 $\le 1$。并且也不能等于 $N-1$， 即 $i=0,j=N-1$ 或 $j=0,i=N-1$ 的情况。

包含 $P_{i+1}$ 的部分的面积求法：如图，$S_{BCDE}=S_{\mathrm{△}BCD}+S_{\mathrm{△}BDE}$，即将多边形以 $P_i$ 为顶点划分为三角形。

三角形面积：已知 $\mathrm{△}ABC$ 的三个顶点 $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$，可以用向量叉乘表示面积： $2S_{\mathrm{△}ABC}=\overrightarrow{AB}\times \overrightarrow{AC}$，即：

$$S_{\mathrm{△}ABC}=\frac{1}{2}\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ x_2-x_1& y_2-y_1& 0\\ x_3-x_1& y_3-y_1& 0\end{array}\right|=\frac{1}{2}|(x_3-x_1)(y_2-y_1)-(x_2-x_1)(y_3-y_1)|$$

代码：

#include <bits/stdc++.h>

using namespace std;

const int SIZE = 1e5 + 10;
int N;
pair<int, int> points[SIZE];

inline long long areax2(pair<int, int> a, pair<int, int> b, pair<int, int> c)
{
    return abs(1ll * (c.first - a.first) * (b.second - a.second) - 1ll * (b.first - a.first) * (c.second - a.second));
}

int main()
{
    cin >> N;
    for (int i = 0; i < N; i++)
        cin >> points[i].first >> points[i].second;
    long long ax8 = 0;
    for (int i = 1; i < N - 1; i++)
    {
        ax8 += areax2(points[0], points[i], points[i + 1]);
    }
    long long ans = INT64_MAX;
    // START
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < N; j++)
        {
            if (abs(i - j) <= 1 || abs(i - j) == N - 1)
                continue;
            long long bx2 = 0;
            for (int k = (i + 1) % N; k != j; k = (k + 1) % N)
            {
                bx2 += areax2(points[i], points[k], points[(k + 1) % N]);
            }
            ans = min(ans, abs(ax8 - 4 * bx2));
        }
    }
    // END
    cout << ans << endl;
    return 0;
}

$O(N^2)$ 法：TLE x11

上面的方法是从多边形分割为小三角形计算，同一个小三角形的面积重复计算了很多次。我们可以用小三角形的面积来组成多边形，可以减少很多次计算。

思路：

在 $P_0,P_1,\cdots ,P_N$ 任取 $P_i$，逆时针方向依次加上以 $P_i$ 为顶点的小三角形的面积，每次加上后计算一次结果的最小值。

如下图，若取 $B$ 为顶点，那么计算 $S_{\mathrm{△}BCD},S_{\mathrm{△}BCD}+S_{\mathrm{△}BDE}$，依此为吃掉的面积计算 $8\times |a-b|$ 的结果。

求法：

取 $j=(i+1)modN$，将已有面积加上以 $P_i,P_j,P_{(j+1)modN}$ 为顶点的三角形的面积，每次操作后将 $j$ 更新为 $(j+1)modN$，一共进行 $N-2$ 次操作（因为不能沿着边切割）

取模很重要！只要是 $+1$ ，序号都有可能超过 $N$ 而重新从 $0$ 开始。

代码：

// 将此代码替换到方法一代码的START注释与END注释之间
for (int i = 0; i < N; i++)
{
    long long bx2 = 0;
    for (int j = (i + 1) % N, t = 0; t < N - 2; j = (j + 1) % N, t++)
    {
        bx2 += areax2(points[i], points[j], points[(j + 1) % N]);
        ans = min(ans, abs(ax8 - 4 * bx2));
    }
}

$O(N)$ 法：AC

思路：

既然题目要求 $8\times |a-b|$ 最小，那么可以用滑动窗口：若 $b<a$，那就是切少了，得多切点；若 $b>a$，那就是切多了，得少切点。

求法：

$i$ 初始为 $0$，$j$ 初始为 $1$，如果 $b<a$，那么加上以 $P_i,P_j,P_{(j+1)modN}$ 为顶点的三角形面积，$j$ 更新为 $(j+1)modN$ 然后再继续加三角形，直到 $b\ge a$。然后再去掉以 $P_i,P_{(i+1)modN},P_j$ 为顶点的三角形面积，$i$ 加一，直到 $b<a$.

代码：

// 将此代码替换到方法一代码的START注释与END注释之间
long long bx2 = 0;
int j = 1;
for (int i = 0; i < N; i++)
{
    while (bx2 * 4 < ax8)
    {
        bx2 += areax2(points[i], points[j], points[(j + 1) % N]);
        j = (j + 1) % N;
        ans = min(ans, abs(ax8 - 4 * bx2));
    }
    bx2 -= areax2(points[i], points[(i + 1) % N], points[j]);
    ans = min(ans, abs(ax8 - 4 * bx2));
}