POJ 2354 Titanic

标签：计算几何

输入好恶心

先把经纬度的时分秒转换成度，再根据东西经，南北纬取正负号。然后根据球上两点距离公式求。

最后注意精度要取大一点

#include<stdio.h>
#include<math.h>
const double pi=acos(-1.0);
const double eps=1e-5;
int main(){
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif // ONLINE_JUDGE
    double s1[3],s2[3];
    int n,a;
    char c1,c2,c3;
    char st[10],s[100];
    for(int i=0;i<9;i++) scanf("%s",s);

    scanf("%lf^%lf'%lf\" %s",&s1[0],&s1[1],&s1[2],st);
    scanf("%s",s);

    double a1,a2,b1,b2;
    a1=s1[0]+s1[1]/60+s1[2]/3600;
    if(st[0]=='S') a1=-a1;

    scanf("%lf^%lf'%lf\" %s.",&s1[0],&s1[1],&s1[2],st);
    //printf("%lf %lf %lf",s1[0],s1[1],s1[2]);
    b1=s1[0]+s1[1]/60+s1[2]/3600;
    if(st[0]=='W') b1=-b1;

    for(int i=0;i<5;i++) scanf("%s",s);
    scanf("%lf^%lf'%lf\" %s",&s1[0],&s1[1],&s1[2],st);
    a2=s1[0]+s1[1]/60+s1[2]/3600;
    if(st[0]=='S') a2=-a2;

    scanf("%s",s);
    scanf("%lf^%lf'%lf\" %s",&s1[0],&s1[1],&s1[2],st);
    b2=s1[0]+s1[1]/60+s1[2]/3600;
    if(st[0]=='W') b2=-b2;
    scanf("%s",s);

    a1=a1*pi/180;a2=a2*pi/180;b1=b1*pi/180;b2=b2*pi/180;
    double theta,ans;
    theta=acos(cos(a1)*cos(a2)*cos(b1-b2)+sin(a1)*sin(a2));
    ans=6875*0.5*theta;
    //printf("%f",theta);
    printf("The distance to the iceberg: %.2f miles.\n",ans);
    if(ans<100-0.005) printf("DANGER!\n");
    return 0;
}

POJ 2354 Titanic

标签：计算几何

原文地址：http://blog.csdn.net/lj94093/article/details/43967207