标签:double for top visit 内存 www 中心 sub ati
http://www.cnblogs.com/xdlwd086/p/5100425.html
这位学长编了java版本的,于是在借鉴学长的思路的基础上,做出了C++的实现,以此分享。
#include <stdio.h> //定义输入/输出函数
#include <stdlib.h> //定义杂项函数及内存分配函数
#include <ctime>
#include <cstdlib>
#include <string>
#include <sstream>
#include <iostream>
#include <fstream>
#include <iomanip>
#include <cmath>
using namespace std;
#define M_PI 3.14159265358979323846
#define DMax 30.0
long double gps[3150][2];
int gps_visit[3150] = {0};
/* The code of GeoDistance function:
Input: Two coordination {Latitude1, Longitude1, Latitude2, Longitude2 } (type:double)
Output: Distance(Unit: m) (type:double)
*/
long double Rad(long double d){
return d * M_PI / 180.0;
}
//经度 longitude 纬度latitude
long double Geodist(long double lon1, long double lat1, long double lon2, long double lat2){
long double radLat1 = Rad(lat1);
long double radLat2 = Rad(lat2);
long double delta_lon = Rad(lon2 - lon1);
long double top_1 = cos(radLat2) * sin(delta_lon);
long double top_2 = cos(radLat1) * sin(radLat2) - sin(radLat1) * cos(radLat2) * cos(delta_lon);
long double top = sqrt(top_1 * top_1 + top_2 * top_2);
long double bottom = sin(radLat1) * sin(radLat2) +cos(radLat1) * cos(radLat2) * cos(delta_lon);
long double delta_sigma = atan2(top, bottom);
long double distance = delta_sigma * 6378137.0;
return distance;
}
//将2007-10-14-GPS.log文件中的GPS数据提取出来,转换之后另存起来
void init1(){
ifstream in1;
ofstream out1;
in1.open("F:\\研一上\\滴滴算法竞赛\\城市计算\\Task\\Data\\task 1-compression\\2007-10-14-GPS.log");
out1.open("F:\\研一上\\滴滴算法竞赛\\城市计算\\Task\\Data\\task 1-compression\\GPS.txt");
for(int i=0;i<3150;i++){
string temp;
getline(in1, temp);
string gps_E = temp.substr(20,10);
string gps_N = temp.substr(33,9);
out1<<gps_E<<" "<<gps_N<<endl;
}
out1.close();
in1.close();
}
void init2(){
ifstream in2;
ofstream out2;
in2.open("F:\\研一上\\滴滴算法竞赛\\城市计算\\Task\\Data\\task 1-compression\\GPS.txt");
out2.open("F:\\研一上\\滴滴算法竞赛\\城市计算\\Task\\Data\\task 1-compression\\realGPS1.txt");
for(int i=0;i<3150;i++){
long double gps_E,gps_N;
in2>>gps_E>>gps_N;
gps_E = (gps_E - 11600.0)*1.0/60+116.0;
gps_N = (gps_N - 3900)*1.0/60+39.0;
out2 <<setiosflags(ios::fixed)<<setprecision(6)<<gps_E<<" "<<gps_N<<" "<<i+1<< endl;
gps[i][0] = gps_E;
gps[i][1] = gps_N;
}
out2.close();
in2.close();
}
long double get_d(int point_A,int point_B,int point_C ){
long double a = abs( Geodist(gps[point_B][0],gps[point_B][1],gps[point_C][0],gps[point_C][1] ) );
long double b = abs( Geodist(gps[point_A][0],gps[point_A][1],gps[point_C][0],gps[point_C][1] ) );
long double c = abs( Geodist(gps[point_A][0],gps[point_A][1],gps[point_B][0],gps[point_B][1] ) );
long double p = (a+b+c)/2.0;
long double s = sqrtl( abs( p*(p-a)*(p-b)*(p-c) ) );
long double d = s*2.0/c;
return d;
}
void dp_gps(int point_start,int point_end){
if(point_start<point_end){ //递归进行条件
long double maxDist = 0; //最大距离
int mid = 0; //最大距离对应的下标
for(int i=point_start+1;i<point_end;i++){
long double temp = get_d(point_start,point_end,i);
if(temp>maxDist){
maxDist = temp;
mid = i;
}//求出最大距离及最大距离对应点的下标
}
if(maxDist>=DMax){
gps_visit[mid] = 1; //记录当前点加入
//将原来的线段以当前点为中心拆成两段,分别进行递归处理
dp_gps(point_start,mid);
dp_gps(mid,point_end);
}
}
}
int main(){
int count = 0; //记录输出点的个数
long double Mean_distance_error; //平均距离误差
long double Compression_rate; //压缩率
init1();
init2();
gps_visit[0] = 1;
gps_visit[3149] = 1;
dp_gps(0,3149);
ofstream out3;
out3.open("F:\\研一上\\滴滴算法竞赛\\城市计算\\Task\\Data\\task 1-compression\\pointID.txt");
for(int i=0;i<3150;i++ ){
if(gps_visit[i]==1){
out3<<i+1<<endl;
count++;
}
}
out3.close();
long double sum_notVisit_d = 0;
int start = 0,end;
for(int i=0;i<3150;){
if(start == 3149) break; //如果开始点是尾点,那就结束
for(int j=start+1;j<3150;j++){ //找出下一个压缩节点
if(gps_visit[j]==1){
end = j;
break;
}
}
for(int k=start+1;k<end;k++ ){
if(gps_visit[k]==0){
sum_notVisit_d+= get_d( start,end,k );
}
}
start = end;
}
Mean_distance_error = sum_notVisit_d/3150.0;
Compression_rate = count/3150.0;
cout<<count<<endl; //输出压缩后点的个数
cout<<setiosflags(ios::fixed)<<setprecision(6)<<Mean_distance_error <<endl;
cout<<setiosflags(ios::fixed)<<setprecision(4)<<Compression_rate*100<<"%"<<endl;
system("pause");
return 0;
}
标签:double for top visit 内存 www 中心 sub ati
原文地址:http://www.cnblogs.com/liugl7/p/7777660.html