http://acm.hdu.edu.cn/diy/contest_showproblem.php?cid=23003&pid=1003
题解：已知经纬度，求两点之间距离。然后最小生成树……
1 // 2 // main.cpp 3 // xiecheng3 4 // 5 // Created by zhang on 14-4-10. 6 // Copyright (c) 2014年 apple. All rights reserved. 7 // 8 9 // 10 // main.cpp 11 // POJ 1258 12 // 13 // Created by zhang on 14-3-31. 14 // Copyright (c) 2014年 apple. All rights reserved. 15 // 16 17 #include <iostream> 18 #include <cstdio> 19 #include <algorithm> 20 #include <cmath> 21 22 using namespace std; 23 24 const int maxn=1100; 25 const int INF=1001000; 26 int cost[maxn][maxn]; 27 int MINC[maxn]; 28 bool used[maxn]; 29 int V; 30 long long loc[maxn]; 31 double lat[maxn],lng[maxn]; 32 const double PI=3.14159265358979323846; 33 double r ; 34 int DIS[maxn]; 35 36 double rad(double d) 37 38 { 39 40 return d * PI / 180.0; 41 42 } 43 44 45 46 double getDistance(double lat1, double lng1, double lat2, double lng2) 47 48 { 49 50 double radLat1 = rad(lat1); 51 52 double radLat2 = rad(lat2); 53 54 double radLng1 = rad(lng1); 55 56 double radLng2 = rad(lng2); 57 58 double s = acos(sin(radLat1)*sin(radLat2)+cos(radLat1)*cos(radLat2)*cos(radLng1-radLng2)); 59 60 s = s * r; 61 62 return s; 63 64 } 65 66 int prim() 67 { 68 for (int i=0; i<V; i++) { 69 MINC[i]=INF; 70 used[i]=false; 71 } 72 MINC[0]=0; 73 int res=0; 74 while (1) { 75 int v=-1; 76 for (int u=0; u<V; u++) { 77 if (!used[u]&&(v==-1||MINC[u]<MINC[v])) { 78 v=u; 79 } 80 } 81 if (v==-1) { 82 break; 83 } 84 used[v]=true; 85 res+=MINC[v]; 86 for (int u=0; u<V; u++) { 87 MINC[u]=min(MINC[u],cost[v][u]); 88 } 89 } 90 return res; 91 } 92 93 94 int main() 95 { 96 //freopen("/Users/apple/Desktop/xiecheng3/xiecheng3/in", "r", stdin); 97 //freopen("/Users/apple/Desktop/xiecheng3/xiecheng3/out", "w", stdout); 98 int N; 99 int L,C; 100 double D; 101 102 103 scanf("%d",&N); 104 for (int i=0; i<N; i++) { 105 D=0; 106 L=0; 107 C=0; 108 scanf("%lf",&D); 109 //printf("%lf ",D); 110 r=D/2; 111 scanf("%d",&L); 112 //printf("%d ",L); 113 scanf("%d",&C); 114 V=C; 115 //printf("%d ",C); 116 for (int i=0; i<C; i++) { 117 cin>>lat[i]>>lng[i]; 118 } 119 for (int i=0; i<C; i++) { 120 for (int j=0; j<C; j++) { 121 cost[i][j]=getDistance(lat[i],lng[i],lat[j],lng[j]); 122 //printf("cost %d %d %d ",cost[i][j],i,j); 123 } 124 } 125 //printf("%d ",prim()); 126 if (prim()>L) { 127 printf("N "); 128 } 129 if (prim()<=L) { 130 printf("Y "); 131 } 132 } 133 return 0; 134 }