题意
就是给你两条线段AB , CD ，一个人在AB以速度p跑,在CD上以q跑,
在其他地方跑速度是r。问你从A到D最少的时间。
三分AB ，然后再三分CD ，模板题目，这题卡精度 eps不能少
1 #include <cstdio>
2 #include <cstring>
3 #include <queue>
4 #include <cmath>
5 #include <algorithm>
6 #include <set>
7 #include <iostream>
8 #include <map>
9 #include <stack>
10 #include <string>
11 #include <vector>
12 #define pi acos(-1.0)
13 #define eps 1e-9
14 #define fi first
15 #define se second
16 #define rtl rt<<1
17 #define rtr rt<<1|1
18 #define bug printf("******\n")
19 #define mem(a,b) memset(a,b,sizeof(a))
20 #define name2str(x) #x
21 #define fuck(x) cout<<#x" = "<<x<<endl
22 #define f(a) a*a
23 #define sf(n) scanf("%d", &n)
24 #define sff(a,b) scanf("%d %d", &a, &b)
25 #define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
26 #define sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
27 #define pf printf
28 #define FRE(i,a,b) for(i = a; i <= b; i++)
29 #define FREE(i,a,b) for(i = a; i >= b; i--)
30 #define FRL(i,a,b) for(i = a; i < b; i++)+
31 #define FRLL(i,a,b) for(i = a; i > b; i--)
32 #define FIN freopen("data.txt","r",stdin)
33 #define gcd(a,b) __gcd(a,b)
34 #define lowbit(x) x&-x
35 using namespace std;
36 typedef long long LL;
37 typedef unsigned long long ULL;
38 const int mod = 1e9 + 7;
39 const int maxn = 2e5 + 10;
40 const int INF = 0x3f3f3f3f;
41 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
42 int t;
43 double P, Q, R, len1, len2;
44 struct node {
45 double x, y;
46 } a, b, c, d, temp1, temp2;
47 double dist ( node a, node b ) {
48 return sqrt ( eps + ( a.x - b.x ) * ( a.x - b.x ) + ( a.y - b.y ) * ( a.y - b.y ) );
49 }
50 void caltemp1 ( double cnt ) {
51 temp1.x = a.x + ( b.x - a.x ) / len1 * cnt;
52 temp1.y = a.y + ( b.y - a.y ) / len1 * cnt;
53 }
54 void caltemp2 ( double cnt ) {
55 temp2.x = d.x + ( c.x - d.x ) / len2 * cnt;
56 temp2.y = d.y + ( c.y - d.y ) / len2 * cnt;
57 }
58 double check ( double cnt ) {
59 caltemp1 ( cnt );
60 double l = 0, r = len2, ll, rr, ans1, ans2;
61 while ( r - l > eps ) {
62 ll = ( 2 * l + r ) / 3, rr = ( 2 * r + l ) / 3;
63 caltemp2 ( ll );
64 ans1 = dist ( temp1, temp2 ) / R + dist ( d, temp2 ) / Q;
65 caltemp2 ( rr );
66 ans2 = dist ( temp1, temp2 ) / R + dist ( d, temp2 ) / Q;
67 if ( ans1 < ans2 ) r = rr;
68 else l = ll;
69 }
70 return ans1 + dist ( temp1, a ) / P;
71 }
72 int main() {
73 scanf ( "%d", &t );
74 while ( t-- ) {
75 scanf ( "%lf%lf%lf%lf", &a.x, &a.y, &b.x, &b.y );
76 scanf ( "%lf%lf%lf%lf", &c.x, &c.y, &d.x, &d.y );
77 scanf ( "%lf%lf%lf", &P, &Q, &R );
78 len1 = dist ( a, b ), len2 = dist ( c, d );
79 double l = 0, r = len1, ll, rr;
80 while ( r - l > eps ) {
81 ll = ( 2 * l + r ) / 3, rr = ( 2 * r + l ) / 3;
82 if ( check ( ll ) < check ( rr ) ) r = rr;
83 else l = ll;
84 }
85 printf ( "%.2f\n", check ( ll ) );
86 }
87 return 0;
88 }
来源：oschina
链接：https://my.oschina.net/u/4265175/blog/3746959