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poj2354Titanic(两点的球面距离)

时间:2014-07-26 01:13:36      阅读:338      评论:0      收藏:0      [点我收藏+]

链接

球面距离计算公式:d(x1,y1,x2,y2)=r*arccos(sin(x1)*sin(x2)+cos(x1)*cos(x2)*cos(y1-y2))
x1,y1是纬度\经度的弧度单位,r为地球半径
 
把有用的信息从恶心的字符串中提取出来,然后套公式,再处理下恶心的精度,得出的距离需要四舍五入。
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 1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 #include<vector>
 7 #include<cmath>
 8 #include<queue>
 9 #include<set>
10 using namespace std;
11 #define N 100000
12 #define LL long long
13 #define INF 0xfffffff
14 const double eps = 1e-8;
15 const double pi = acos(-1.0);
16 const double inf = ~0u>>2;
17 const double r = 6875/2.0;
18 char s[110];
19 int dcmp(double x)
20 {
21     if(fabs(x)<eps) return 0;
22     return x<0?-1:1;
23 }
24 void solve(double x1,double y1,double x2,double y2)
25 {
26     double ans =  r*acos(sin(x1)*sin(x2)+
27                    cos(x1)*cos(x2)*cos(y1-y2));
28 
29     printf("The distance to the iceberg: %.2lf miles.\n",ans);
30     if(dcmp(ans-100+0.005)<0)
31     puts("DANGER!");
32 }
33 int main()
34 {
35     double d1,d2,d3,d4;
36     int a,b,c,i;
37     char dir[10];
38     for(i = 0; i < 9 ; i++)
39     scanf("%s",s);
40     scanf("%s",s);
41     sscanf(s,"%d^%d‘%d‘‘",&a,&b,&c);
42     //printf("%d %d %d\n",a,b,c);
43     d1 = a+b/60.0+c/3600.0;
44     scanf("%s",dir);
45     if(dir[0]==S)
46     d1=-d1;
47     scanf("%s",s);
48 
49     scanf("%s",s);
50     sscanf(s,"%d^%d‘%d‘‘",&a,&b,&c);
51     //printf("%d %d %d\n",a,b,c);
52     d2 = a+b/60.0+c/3600.0;
53     scanf("%s",dir);
54     if(dir[0]==W)
55     d2=-d2;
56 
57     for(i =0 ; i < 5 ; i++) scanf("%s",s);
58     scanf("%s",s);
59     sscanf(s,"%d^%d‘%d‘‘",&a,&b,&c);
60     //printf("%d %d %d\n",a,b,c);
61     d3 = a+b/60.0+c/3600.0;
62     scanf("%s",dir);
63     if(dir[0]==S)
64     d3=-d3;
65 
66     scanf("%s",s);
67 
68     scanf("%s",s);
69     sscanf(s,"%d^%d‘%d‘‘",&a,&b,&c);
70     //printf("%d %d %d\n",a,b,c);
71     d4 = a+b/60.0+c/3600.0;
72     scanf("%s",dir);
73     if(dir[0]==W)
74     d4=-d4;
75     scanf("%s",s);
76 
77     solve(d1*pi/180,d2*pi/180,d3*pi/180,d4*pi/180);
78     return 0;
79 }
View Code

 

poj2354Titanic(两点的球面距离),布布扣,bubuko.com

poj2354Titanic(两点的球面距离)

原文:http://www.cnblogs.com/shangyu/p/3868819.html

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