HDU 3400 Line belt 【三分嵌套】

【5G核心网】Packet Forwarding Model 数据包转发模型 CP 通过建立,修改,删除 PFCP 会话上下文,以及通过为每个 PFCP 会话赏析哪问提供 PDR FAR QER URR BAR,来控制 UP 面的数据包处理。 每一个 PDR 都包含 PDI(Packet Detection Information,包探测信息),一个或多个字段,与输入数据包进行匹配,并可能与以下规则相关联,从而提供了适用于与 PDI 匹配的数据包的指令集: - One FAR(包含了处理数据包的相关指令) an Apply Act... 阅读详情

 

In a two-dimensional plane there are two line belts, there are two segments AB and CD, lxhgww's speed on AB is P and on CD is Q, he can move with the speed R on other area on the plane. 
How long must he take to travel from A to D?

InputThe first line is the case number T. 
For each case, there are three lines. 
The first line, four integers, the coordinates of A and B: Ax Ay Bx By. 
The second line , four integers, the coordinates of C and D:Cx Cy Dx Dy. 
The third line, three integers, P Q R. 
0<= Ax,Ay,Bx,By,Cx,Cy,Dx,Dy<=1000 
1<=P,Q,R<=10OutputThe minimum time to travel from A to D, round to two decimals.Sample Input

1
0 0 0 100
100 0 100 100
2 2 1

Sample Output

136.60

 

 

 

题意:

在AB,CD两条线段,求从A点到D点的最小距离,原先是思路是通过三分的方法找出两条线段对应的 0 <= 

 λ <= 1  使得所走的距离最短;但是,在三分的过程中不知道怎么去处理当
 λ=0的情况,如果等于零的话,所求的线段长度是不准确的,这里可以三分去找出AB内的点和CD内的点,通过点与点直接求线段的距离,用struct来存储x和y的坐标,这里的三分嵌套就是,先给AB三分,然后在AB的基础上再给CD三分,下面给出代码:

 

 

#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<time.h>
#include<algorithm>
#include<cmath>
#include<stack>
#include<list>
#include<queue>
#include<map>
#define E exp(1)
#define PI acos(-1)
#define mod (ll)(1e9+9)
#define INF 0x3f3f3f3f;
#define MAX 40000
#define compare 0.00000001
//#define _CRT_SECURE_NO_WARNINGS
//#define LOCAL
using namespace std;

typedef long long ll;
typedef long double lb;

double Ax, Bx, Ay, By, Cx, Cy, Dx, Dy, p, q, r;
struct point {
	double x;
	double y;
};

struct point A, B, C, D;
   //m 为AB内的点的坐标,mm为CD内的点的坐标
double formula(struct point mm,struct point m) {
	double ans_1 = sqrt(pow(D.x - mm.x, 2) + pow(D.y - mm.y, 2)) / q;
	double ans_2 = sqrt(pow(mm.x - m.x, 2) + pow(mm.y - m.y, 2)) / r;
	double ans_3 = sqrt(pow(A.x - m.x, 2) + pow(A.y - m.y, 2)) / p;
	return ans_1 + ans_2 + ans_3;
}
		//m 为AB内的点的坐标
double devide(struct point m) {
	struct point lle, rri, llmid, rrmid;
	lle = C, rri = D;
	while (sqrt(pow(lle.x - rri.x, 2) + pow(lle.y - rri.y, 2)) > 1e-8) {
		llmid.x = (lle.x + rri.x) / 2;
		llmid.y = (lle.y + rri.y) / 2;
		rrmid.x = (llmid.x + rri.x) / 2;
		rrmid.y = (llmid.y + rri.y) / 2;
		if (formula(llmid,m) > formula(rrmid,m)) lle = llmid;
		else rri = rrmid;
	}
	return formula(rri, m);
}

int main(void)
{
#ifdef LOCAL
	freopen("data.in.txt", "r", stdin);
	freopen("data.out.txt", "w", stdout);
#endif
	int t;
	while (scanf("%d", &t) != EOF) {
		while (t--) {
			scanf("%lf%lf%lf%lf", &A.x, &A.y, &B.x, &B.y);
			scanf("%lf%lf%lf%lf", &C.x, &C.y, &D.x, &D.y);
			scanf("%lf%lf%lf", &p, &q, &r);

			struct point le, ri, lmid, rmid;
			le = A, ri = B;
			while (sqrt(pow(le.x - ri.x, 2) + pow(le.y - ri.y, 2)) > 1e-8) {
				lmid.x = (le.x + ri.x) / 2;
				lmid.y = (le.y + ri.y) / 2;
				rmid.x = (lmid.x + ri.x) / 2;
				rmid.y = (lmid.y + ri.y) / 2;
				if (devide(lmid) > devide(rmid)) le = lmid;
				else ri = rmid;
			}
			double res = devide(ri);
			printf("%.2f\n", res);
		}
	}
	//	end = clock();
	//	cout << "using tmie:" << (double)(end - start) / CLOCKS_PER_SEC * (1000) << "ms" << endl;

	//system("pause");
	return 0;
}

 

Freeswitch 如何检测链路是否有回铃音 Freeswitch回铃音检测解决方案 问题描述:呼叫手机时,虽然信令正常(183/180),但部分手机无回铃音,导致坐席无法判断是否接通,用户体验差。 解决方案: 先播放本地音频作为回铃音 实时检测用户侧是否有声音输入 检测到声音后自动切换为用户侧音频 技术实现: 使用Freeswitch APP实现功能 创建结构体保存实例数据(包括铃声文件、会话句柄、VAD检测器等) 通过media bug机制实现音频流的替换 利用VAD(语音活动检测)技术监控用户侧声音 当检测到用户说话时自动终止本地铃声播放 该方案 阅读详情

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主要解决vnc-viewer灰屏,黑屏,鼠标大黑叉等现象,详细见文档。

Line belt HDU - 3400嵌套三分

嵌套三分 #include &lt;iostream&gt; #include &lt;cmath&gt; #include &lt;cstdio&gt; #include &lt;algorithm&gt; using namespace std; const double eps=1e-9; struct Point { double x,y; Point(){} ...

leekerian的博客 184

HDU 3400 Line belt(嵌套三分)

Description In a two-dimensional plane there are two line belts, there are two segments AB and CD, lxhgww's speed on AB is P and on CD is Q, he can move with the speed R on other area on the plane.

千里之行 始于足下 749

HDU/HDOJ 3400 Line belt 三分嵌套 杭电月赛

Line belt Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Total Submission(s) : 24   Accepted Submiss

bingshen的专栏 1168

Line belt 三分嵌套

In a two-dimensional plane there are two line belts, there are two segments AB and CD, lxhgww's speed on AB is P and on CD is Q, he can move with the speed R on other area on the plane. How long must h

zero 339

hdu 3400 Line belt

Line belt                                                                        Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)                                

ACway的专栏 3074

Line belt(HDU 3400)(三分套三分)

思路瞅的某dalao的这篇博客 https://www.cnblogs.com/cjjsb/p/9179699.html /* 三分套三分:AB边存在E点,CD边存在F点,假设E点已经在了,只需确定F点的话,可以知道F点在CD边上移动是一个凹形函数,F点已知,确定E点的话 E点在AB上移动其距离函数形成的曲线是一个直线,直线+凹形函数也是凹形函数,所以可以用三分做。在三分确定E的时候,假装E点...

Autumn_ghost的博客 304

三分套三分 --- HDU 3400 Line belt

 Line belt Problem's Link:   http://acm.hdu.edu.cn/showproblem.php?pid=3400   Mean:  给出两条平行的线段AB, CD,然后一个人在线段AB的A点出发,走向D点,其中,人在线段AB上的速度为P, 在线段CD上的速度为Q,在其他地方的速度为R,求人从A点到D点的最短时间。   analyse: 经典的...

weixin_34365417的博客 163

HDU 3400 Line belt

题目大意 给出两条平行的线段AB, CD,然后一个人在线段AB的A点出发,走向D点,其中,人在线段AB上的速度为P, 在线段CD上的速度为Q,在其他地方的速度为R,求人从A点到D点的最短时间。 解题思路 如上图所示,红线部分为人走的路径。人所用的时间为 T = X / P + Y / Q + Z / R。 然后我们做一个变形,令人在线段AB上花的时间为:F(X) = X / P ,

DSL_HN_2002的博客 515

HDU 3400 Line belt(三分)

这道题是典型的三分,因为从一点到另一条直线上的点的距离的变化不是单调的,而是先减后增,所以需要用嵌套三分。首先三分AB,再依次以mid和mmid为起点,CD上的某点为终点三分即可。

Noooooorth的专栏 1743

hdu3400 Line belt

如图所示,必然在AB,CD上存在两点e,f使A到D的时间最小,由此可以假设f点已知,由此确定下e点,所以采用嵌套三分 code /* ID: yueqiq PROG: numtri LANG: C++ */ #include #include #include #include #include #include #include #include #include #inc

yueqiq 521
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