UVa 11722 - Joining with Friend （概率 数形结合）

 

I I U P C   2 0 0 9
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 midnight00:00.
Output
For each test case print one line of output in the format “Case #k: p” Here k is the case number andp 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

 

题意：

你乘火车Ａ　你的小伙伴乘火车Ｂ，你们想要在Ｘ站会面

火车A到Ｘ站的时间可能是[t1, t2]中的某时刻（等概率），火车Ｂ到Ｘ站的时间可能是[s1, s2]中的某时刻（等概率）。

ＡＢ车到站后都会停留Ｗ分钟。只有在同一时刻ＡＢ车都停在Ｘ站的时候，你才可以和你小伙伴会面。

求你们能够会面的概率

大白上的思路很巧妙　把问题转化为求面积

ｘ轴表示Ａ车到站的时刻　即[t1, t2]是一条平行于ｘ轴的线段L1

ｙ轴表示Ｂ车到站的时刻　即[s1, s2]是一条平行于ｙ轴的线段L2

则L1 L2就是一个矩形的两条相邻的边，整个概率空间就是这个矩形

则会面的概率就是　矩形与（直线y = x + w 与 y = x - w之间的面积）的相交部分

#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
#include <queue>
#include <set>

using namespace std;

//#define WIN
#ifdef WIN
typedef __int64 LL;
#define iform "%I64d"
#define oform "%I64d\n"
#define oform1 "%I64d"
#else
typedef long long LL;
#define iform "%lld"
#define oform "%lld\n"
#define oform1 "%lld"
#endif

#define S64I(a) scanf(iform, &(a))
#define P64I(a) printf(oform, (a))
#define P64I1(a) printf(oform1, (a))
#define REP(i, n) for(int (i)=0; (i)<n; (i)++)
#define REP1(i, n) for(int (i)=1; (i)<=(n); (i)++)
#define FOR(i, s, t) for(int (i)=(s); (i)<=(t); (i)++)

const int INF = 0x3f3f3f3f;
const double eps = 1e-9;
const double PI = (4.0*atan(1.0));

double t1, t2, s1, s2, w;
double all_s;

double area(double d) {
    if(t1 + d > s2-eps) return all_s;
    if(t2 + d < s1+eps) return 0;
    double ret = 0;
    double x1 = s1 - d, x2 = t2 + d;
    double x3 = t1 + d, x4 = s2 - d;
    if(t1-eps < x1 && x1 < t2 + eps) {
        if(s1-eps < x2 && x2 < s2 + eps) {
            ret = (t2-x1) * (x2-s1) * 0.5;
        } else {
            ret = (t2-x1 + t2-x4) * (s2-s1) * 0.5;
        }
    } else {
        if(s1-eps < x2 && x2 < s2 + eps) {
            ret = (x3-s1 + x2-s1) * (t2-t1) * 0.5;
        } else {
            ret = all_s - (s2-x3)*(x4-t1) * 0.5;
        }
    }
    return ret;
}

int main() {
    int T;

    scanf("%d", &T);
    for(int kase=1; kase<=T; kase++) {
        scanf("%lf%lf%lf%lf%lf", &t1, &t2, &s1, &s2, &w);
        all_s = (t2 - t1) * (s2 - s1);
        double s1 = area(w);
        double s2 = area(-1*w);
        double ans = (s1 - s2) / all_s;
        printf("Case #%d: %.8lf\n", kase, ans);
    }

    return 0;
}