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给出每个点的经纬，和航线（有向边），求最短路
题意很简单，主要是距离的转化，先把经纬度转化成弧度数，然后转化出每个点的（x，y，z）然后在球上求距离。
#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <map>#include <cmath>#include <queue>#include <cmath>#include <stack>#include <stdio.h>#include <string.h>#include <math.h>#include <iostream>#define INF  0x3f3f3f3fusing namespace std;int n, m,q ;const double PI = 3.141592653589793;const double r = 6378 ;const int maxn = 10000+10;struct Point{    double lt,ln;    double ht,hn;    string name;    double x,y,z;} s[maxn];map<string,int>ma;int E;int g[200][200];int get_id(string s1){    if(ma[s1])    {        return ma[s1];    }    else    {        E++;        ma[s1] = E;        return E;    }}int get_dis(int i,int j){    double l = acos((s[i].x * s[j].x + s[i].y * s[j].y  + s[i].z * s[j].z ) / (r*r));    l = round(l * r);        return l;}void get_pos(int i){    double l = r * cos(s[i].ht);    s[i].x = l * cos(s[i].hn);    s[i].z = r * sin(s[i].ht);    s[i].y = l * sin(s[i].hn);//    cout << "*******" << endl;//    printf("%lf %lf %dlf\n",s[i].x,s[i].y,s[i].z);//    cout << "*******" << endl;}void floyd(){    for(int k = 1 ; k <= E ; k++)    {        for(int i = 1 ; i <= E ; i++)        {            for(int j = 1 ; j <= E ; j++)            {                if(g[k][j] != INF && g[i][k] != INF)                    g[i][j] = min(g[i][j],g[i][k] + g[k][j]);            }        }    }}int main(){    int kase = 1;    while(cin >> n >> m >> q && (n||m||q))    {        E = 0;        ma.clear();        memset(g,0,sizeof(g));        for(int i = 0 ; i < n ; i++)        {            cin >> s[i].name;            scanf("%lf%lf",&s[i].lt,&s[i].ln);            s[i].ht = s[i].lt * PI / 180;            s[i].hn = s[i].ln * PI / 180;            get_pos(i);        }        for(int i = 0 ; i <= n ; i++)        {            for(int j = 0 ; j <= n ; j++)            {                if(i!=j)                {                    g[i][j] = INF;                }            }        }//        for(int i = 0 ; i < n ; i++)//        {//            for(int j = 0 ; j < n ;j++)//            {////                int from =//                if(i!=j)//                {//                    g[get_id(s[i].name)][get_id(s[j].name)] = get_dis(i,j);//                }//            }//        }        string s1,s2;        for(int i = 0 ; i< m ; i++)        {            cin >> s1 >> s2;            int x1 = get_id(s1);            int x2 = get_id(s2);            bool flag = false;            for(int j = 0; j < n ; j++)            {                for(int k = 0 ; k < n ; k++)                {                    if(s[j].name == s1 && s[k].name == s2)                    {                        g[x1][x2] = get_dis(j,k);//                        cout << "get:" << get_dis(j,k) << endl;                        flag=true;                        break;                    }                }                if(flag) break;            }        }        floyd();//        cout << "E : " << endl;//        for(int k = 1 ; k <= E ; k++)//        {//            for(int i = 1 ; i <= E ; i++)//            {//                g[i][j] = min(g[i][j],g[i][k] + g[k][j]);//                }//            }//        }        if(kase > 1) cout <<endl;        printf("Case #%d\n",kase++);        for(int i = 0 ; i< q ; i++)        {            cin >> s1 >> s2;            int x1 = get_id(s1);            int x2 = get_id(s2);//            cout << "xxxx" << x1 << " " << x2 << endl;//            for(int k = 1 ; k <= E ; k++)//        {//            for(int i = 1 ; i <= E ; i++)//            {//            if(g[x1][x2] != INF)            {                printf("%d km\n",g[x1][x2]);            }            else            {                printf("no route exists\n");            }        }//        if()    }    return 0;}
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