UVA11722概率问题之线性规划

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 following T line will contain 5 integers t1, t2, s1, s2, w (360 ≤ t1 < t2 < 1080, 360 ≤ s1 < s2 < 1080 and 1 ≤ w ≤ 90). All inputs t1, t2, s1, s2 and w are given in minutes and t1, t2, s1, s2 are minutes since midnight 00:00.
Output
For each test case print one line of output in the format “Case #k: p” Here k is the case number and p is the probability of seeing your friend. Up to 1e-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
",cas,ans);
    }
    return 0;
}