Cycle Function - 题解【二分】

题面

这是2019四川省赛的C题。这里放上CodeForces的链接：C. Cycle Function - Codeforces

Fish is learning functions! He has a linear function $f(x)=Ax+B$ and $N$ numbers $x_1,x_2,⋯,x_N$. Now he is curious about for each function $g(x)$ in

$$\left\{ \begin{matrix} g_1(x)=c_1x+d_1\\ g_2(x)=c_2x+d_2\\ \vdots\\ g_M(x)=c_Mx+d_M \end{matrix} \right\}$$

how to calculate the difference between $f(g(x))$ and $g(f(x))$.

As smart as Fish is, he soon comes up with a function $D(x)=|f(g(x))−x|+|g(f(x))−x|$ and uses the sum over $x_1,x_2,⋯,x_N$ as the difference.

Can you tell him all the differences immediately?

Input
The first line of input contains an integer $T$, representing the number of test cases.

Then for each test case:

The first line contains two integers $N, M$ as mentioned above and then two real numbers $A, B$ indicating the given function $f(x)=Ax+B$.

The second line contains $N$ real numbers $x_1,x_2,⋯,x_N$.

Then $M$ lines follow, each line containing two real numbers $c_i, d_i$ indicating a function $g_i(x)=c_ix+d_i$ mentioned above.

All numbers in the same line are separated by one space.

Output
For each test case, you should output Case x: in the first line, where $x$ indicates the case number starting from $1$.

Then $M$ lines follow, the $i$-th line of which contains a real number representing the difference for given function $g_i(x)$.

Your answers will be considered correct if its absolute error does not exceed $10^{−6}$.

Limits

time limit per test: 5 seconds
memory limit per test: 256 megabytes
input: standard input
output: standard output

Note

$1≤T≤100$
$1≤N,M≤10^5$
$−100≤A,B,x_i,c_i,d_i≤100$
For $90\%$ test cases: $max(N,M)≤1000$

Example Input

2
3 2 2.0 3.0
1.0 2.0 3.0
0.4 -2.0
0.6 -5.0
3 2 2.5 2.0
1.0 2.0 3.0
0.4 -2.0
0.6 -5.0

Example Output

Case 1:
7.800000
28.200000
Case 2:
12.600000
36.900000

大意

给定一个线性函数$f(x)=Ax+B$和一个线性函数组$g_i(x)=c_ix+d_i$，定义函数$D(x)=|f(g(x))−x|+|g(f(x))−x|$，对于给定的一组数$x_i,\enspace i=i,2,\dots,n$，求关于线性函数组里的每个函数，$\sum_{i=1}^{n} D(x_i)$的值。

题解

我们对这两个绝对值符号里面的内容做一下数学变换，记$Ac-1=a,Ad+B=p,Bc+d=q$：

$$f(g(x))-x=A(cx+d)+B=Acx+Ad+B-x=(Ac-1)x+(Ad+B)=ax+p$$

$$g(f(x))-x=c(Ax+B)+d=Acx+Bc+d-x=(Ac-1)x+(Bc+d)=ax+q$$

式子变得简单了不少。但是绝对值仍然是一个让人非常讨厌的东西，需要对正数与负数进行分类讨论。而且，本题有至多1e5个变量和 1e5个函数，虽然时间限制是5秒，但是平方复杂度的暴力解法必定会超时。我们考虑到，对于线性函数$F(x)=kx+b$，当$k=0$时值永远为$b$，否则其零点为$x_0=-\frac{b}{k}$，因此我们只需要特判系数$a$是否为0，然后考虑大于$x_0$的数以及不大于$x_0$的数，分别计算即可，一连串数据的求和可以使用前缀和来维护，分界点排序后使用二分查找函数lower_bound即可。这题知道思路了就很简单，但是确实蛮难想的……

排序复杂度为$O(NlogN)$，对每个函数二分查找总复杂度为$O(MlogN)$，综合复杂度为$O((M+N)logN)$，可以接受。

另：本题卡输入输出，要使用C风格的输入输出或者手搓快读快写。Java或者Python选手还是放弃吧o(￣▽￣)ブ笔者使用关闭流同步的cin\cout超时，改scanf\printf只用了2000ms……

下面是代码：

#include <bits/stdc++.h>
#define GRP int T;cin>>T;rep(C,1,T)
#define FAST ios::sync_with_stdio(false);cin.tie(0);
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define rrep(i,a,b) for(int i=a;i>=b;--i)
#define elif else if
#define mem(arr,val) memset(arr,val,sizeof(arr))
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

const double EPS = 1e-8;
int n, m;
double a, b;
double xs[100010], x[100010], c, d;
double ans;

int main() {
	xs[0] = 0;
	GRP {
		scanf("%d %d %lf %lf", &n, &m, &a, &b);
		printf("Case %d:\n", C);
		//读入并排序，求前缀和
		rep(i, 1, n) {
			scanf("%lf", x + i);
		}
		sort(x + 1, x + 1 + n);
		rep(i, 1, n) {
			xs[i] = xs[i - 1] + x[i];
		}
		//处理每个g(x)
		rep(i, 1, m) {
			scanf("%lf %lf", &c, &d);
			ans = 0;
			double a1 = a * c - 1, p = a * d + b, q = b * c + d;	//计算系数
			if (abs(a1) < EPS) {					//浮点数为0应判断其绝对值小于一个极小值
				ans = (abs(p) + abs(q)) * n;
			} else {
				//寻找分界点
				int i1 = lower_bound(x + 1, x + 1 + n, -p / a1) - x - 1;
				int i2 = lower_bound(x + 1, x + 1 + n, -q / a1) - x - 1;
				//分别求值并且累加
				ans += abs(a1 * (xs[i1] - xs[0]) + p * (i1));
				ans += abs(a1 * (xs[n] - xs[i1]) + p * (n - i1));
				ans += abs(a1 * (xs[i2] - xs[0]) + q * (i2));
				ans += abs(a1 * (xs[n] - xs[i2]) + q * (n - i2));
			}
			printf("%.7lf\n", ans);
		}
	}
	return 0;
}

/*
          _           _    _            _
    /\   | |         | |  | |          (_)
   /  \  | | _____  _| |__| | ___  _ __ _ _ __   __ _
  / /\ \ | |/ _ \ \/ /  __  |/ _ \| '__| | '_ \ / _` |
 / ____ \| |  __/>  <| |  | | (_) | |  | | | | | (_| |
/_/    \_\_|\___/_/\_\_|  |_|\___/|_|  |_|_| |_|\__, |
                                                 __/ |
                                                |___/
*/