Codeforces 1079D Barcelonian Distance（计算几何）

Posted 2021-01-21 ehanla

题目链接：Barcelonian Distance

题意：给定方格坐标，方格坐标上有两个点A，B和一条直线。规定：直线上沿直线走，否则沿方格走。求A到B的最短距离。

题解：通过直线到达的：A、B两点都有两种方式到直线上，最多4种情况，每种情况求出A、B点到直线的距离和直线上新的两点间距离，取4种情况中最优的。

不通过直线到达：$abs(x1-x2)+abs(y1-y2)$，最后与通过直线到达的最优情况比较，得到最优解。

 1 #include <cmath>
 2 #include <cstdio>
 3 #include <algorithm>
 4 using namespace std;
 5 
 6 double a,b,c,ans;
 7 double x1,y1,x2,y2;
 8 
 9 double calx(double y){
10     return (-c-b*y)/a;
11 }
12 
13 double caly(double x){
14     return (-c-a*x)/b;
15 }
16 
17 double cal1(double x1,double x2){
18     double X1,X2,Y1,Y2;
19     double res=0;
20     X1=x1;Y1=caly(x1);
21     res+=abs(Y1-y1);
22     X2=x2;Y2=caly(x2);
23     res+=abs(Y2-y2);
24     res+=sqrt((X1-X2)*(X1-X2)+(Y1-Y2)*(Y1-Y2));
25     return res;
26 }
27 
28 double cal2(double x1,double y2){
29     double X1,X2,Y1,Y2;
30     double res=0;
31     X1=x1;Y1=caly(x1);
32     res+=abs(Y1-y1);
33     X2=calx(y2);Y2=y2;
34     res+=abs(X2-x2);
35     res+=sqrt((X1-X2)*(X1-X2)+(Y1-Y2)*(Y1-Y2));
36     return res;
37 }
38 
39 double cal3(double y1,double x2){
40     double X1,X2,Y1,Y2;
41     double res=0;
42     X1=calx(y1);Y1=y1;
43     res+=abs(X1-x1);
44     X2=x2;Y2=caly(x2);
45     res+=abs(Y2-y2);
46     res+=sqrt((X1-X2)*(X1-X2)+(Y1-Y2)*(Y1-Y2));
47     return res;
48 }
49 
50 
51 double cal4(double y1,double y2){
52     double X1,X2,Y1,Y2;
53     double res=0;
54     X1=calx(y1);Y1=y1;
55     res+=abs(X1-x1);
56     X2=calx(y2);Y2=y2;
57     res+=abs(X2-x2);
58     res+=sqrt((X1-X2)*(X1-X2)+(Y1-Y2)*(Y1-Y2));
59     return res;
60 }
61 
62 int main(){
63     scanf("%lf%lf%lf",&a,&b,&c);
64     scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
65     double ans=abs(x1-x2)+abs(y1-y2);
66     if(a==0||b==0){
67         printf("%.10f
",abs(x1-x2)+abs(y1-y2));
68         return 0;
69     }
70     ans=min(ans,cal1(x1,x2));
71     ans=min(ans,cal2(x1,y2));
72     ans=min(ans,cal3(y1,x2));
73     ans=min(ans,cal4(y1,y2));
74     printf("%.10f
",ans);
75     return 0;
76 }

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