UVA11722概率问题之线性规划
来源:互联网 发布:蒙顶甘露 知乎 编辑:程序博客网 时间:2024/05/16 00:54
链接:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=&problem=2769&mosmsg=Submission+received+with+ID+13959080
Problem J: Joining with Friend
You are going from Dhaka to Chittagong by train and you came to know one of your old friends is going from city Chittagong to Sylhet. You also know that both the trains will have a stoppage at junction Akhaura at almost same time. You wanted to see your friend there. But the system of the country is not that good. The times of reaching to Akhaura for both trains are not fixed. In fact your train can reach in any time within the interval[t1, t2] with equal probability. The other one will reach in any time within the interval[s1, s2] with equal probability. Each of the trains will stop for w minutes after reaching the junction. You can only see your friend, if in some time both of the trains is present in the station. Find the probability that you can see your friend.
Input
The first line of input will denote the number of cases T (T < 500). Each of the followingT line will contain 5 integers t1, t2, s1, s2, w (360 ≤ t1 < t2 < 1080, 360 ≤ s1 < s2 < 1080 and 1 ≤ w ≤ 90). All inputst1, t2, s1, s2 andw are given in minutes and t1, t2, s1, s2 are minutes since midnight00:00.
Output
For each test case print one line of output in the format “Case #k: p” Herek is the case number and p is the probability of seeing your friend. Up to1e-6 error in your output will be acceptable.
Sample Input
Output for Sample Input
2
1000 1040 1000 1040 20
720 750 730 760 16
Case #1: 0.75000000
Case #2: 0.67111111
Problem Setter: Md. Towhidul Islam Talukder
Special Thanks: Samee Zahur, Mahbubul Hasan
简单二维线性规划问题:
输入:5 integers t1, t2, s1, s2, w (360 ≤ t1 < t2 < 1080, 360 ≤ s1 < s2 < 1080 and 1 ≤ w ≤ 90).
目标函数:0<=|s-t|<=w;计算目标的补集更简单
步骤:画出坐标轴,注意分类讨论;
分类讨论:(图形见书P141)
确定4个交点:
A(t1,t1+w),B(s2-w,s2),C(t2,t2-w),D(s1+w,s1)
aim1=0.5*(s2-ay)*(bx-t1),ay>=s1 && bx<=t2;
aim1=0.5*((s2-ay)+(s2-(t2+w))*(t2-t1), ay>=s1 && bx>t2;
aim1=0.5*(s1-w-t1+(bx-t1))*(s2-s1),ay<s1 && bx<=t2;
aim1=0;//否则
aim2=0.5*(s2-ay)*(bx-t1),cy<=s2 && dx>=t1;
aim2=0.5*((t1-w-s1)+(cy-s1))*(t2-t1), cy<=s2 && dx<t1;
aim2=0.5*(t2-(s2+w)+(t2-dx))*(s2-s1),cy>s2 && dx<=t1;
aim2=0;//否则
aim=((s2-s1)*(t2-t1)-aim1-aim2)/(s2-s1)*(t2-t1);
*/
#include <iostream>#include <cmath>#include <algorithm>#include <cstring>#include<cstdio>using namespace std;double t1,t2,s1,s2,w;double area(double b) // 求y=x+b下方在矩形中截取的面积{ double s=(t2-t1)*(s2-s1); double x1=t1,y1=t1+b; double x2=t2,y2=t2+b; if(y2<=s1) // 直线交于矩形右下顶点或者以下 return 0; if(y1<=s1) // 直线交于矩形下面边 { if(y2<=s2) // 直线交于矩形右面边 return 0.5*(y2-s1)*(t2-(s1-b)); else // 直线交于矩形上面边 return 0.5*(t2-s1+b+t2-s2+b)*(s2-s1); } else if(y1<s2) // 直线交于矩形左面边 { if(y2<=s2) //直线交于矩形右面边 return 0.5*(t1+b-s1+t2+b-s1)*(t2-t1); else // 直线交于矩形上面边 return s-0.5*(s2-t1-b)*(s2-t1-b); } else return s;}int main(){ int t; scanf("%d",&t); for(int cas=1;cas<=t;cas++) { scanf("%lf%lf%lf%lf%lf",&t1,&t2,&s1,&s2,&w); double ans=area(w)-area(-w); ans/=(s2-s1)*(t2-t1); printf("Case #%d: %.8lf\n",cas,ans); } return 0;}
- UVA11722概率问题之线性规划
- uva11722(概率)
- UVA11722(见面概率)
- Uva11722 joining with friend 概率 几何计算
- 线性规划问题之MATLAB实现
- 线性规划之飞机航班调度问题
- 算法实验之线性规划解决配料问题
- 线性规划与网络流之7 试题库问题 (网络流)
- 线性规划与网络流24题之运输问题
- 网络流与线性规划24题 之 餐巾计划问题
- 线性规划与网络流24题之 圆桌问题
- 线性规划与网络流24题之 魔术球问题
- 线性规划与网络流24题之 试题库问题
- 关于matlab 线性规划问题
- matlab求解线性规划问题
- 线性规划求解路径问题
- matlab 求解线性规划问题
- JZOJ5259. 线性规划问题
- Hibernate中离线查询DetachedCriteria案例
- Servlet基础
- hibernate-Query的list方法与iterator方法的区别
- 微波射频工程师必读经典参考书
- 【夯实基础】JSP的9个内置对象
- UVA11722概率问题之线性规划
- Cstyle的札记,Freertos内核详解,第1篇
- 暑假第二十一天,7月27号
- define与typedef 区别
- H组sdj政府将自动控制顾客雪佛兰小凤凰路
- 饭店睡过多少个粉丝的风格斯蒂芬
- 文件上传以及限制类型
- 杭电 1284
- 暑假第二十二天,7月28日