這道題是Geohash得模板題,至於什麼是Geohash,網上不管是博客,還是維基百科,解釋都夠詳細了,當然還有這道題的提示裏也解釋的很詳細了(比較推薦看題目裏的解釋,不但詳細,還有僞碼模板),大致就是用經緯度描述地點時,將經緯度通過二分轉化爲爲二進制編碼,然後按先經度後維度,再經度再維度........的順序將兩二進制編碼合併,接着每5位給出其對應的十進制數,再通過base32表得出對應的Geohash。這道題就是完全不用動腦子的套模板啦。
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
#define clegh 10//geohash編碼長度
char Base32[32]= {'0','1','2','3','4','5','6','7','8','9','b',
'c','d','e','f','g','h','j','k','m','n','p',
'q','r','s','t','u','v','w','x','y','z'
};//Base32編碼
char geohash[15]= {'\0'};
void encode(double lat,double lot,int prec)//譯碼
{
memset(geohash,'\0',sizeof(geohash));
double latitude[2] = {-90, 90};
double logitude[2] = {-180, 180};
int length = prec * 5; // 需要的二進制編碼長度
int bits = 0 ;// 記錄二進制碼
int k=0;
for(int i=0; i<length; i++)
{
if(i%2==0)//從0開始偶數位爲經度
{
double mid = (logitude[0] + logitude[1]) / 2;
if(lot > mid)
{
bits=bits*2+1;
logitude[0] = mid;
}
else
{
bits=bits*2;
logitude[1] = mid;
}
}
else
{
double mid = (latitude[0] + latitude[1]) / 2;
if(lat > mid)
{
bits=bits*2+1;
latitude[0] = mid;
}
else
{
bits=bits*2;
latitude[1] = mid;
}
}
if(!((i+1)%5))
{
geohash[k++] = Base32[bits];
//printf("%d\n",bits);
bits = 0;// 重置二進制碼
}
}
}
double lat=0, lot=0;
void docode(char *geoh)//解碼
{
bool odd = true ;// 當前計算位的奇偶性
double latitude[2] = {-90, 90};
double longitude[2] = {-180, 180};
for(int i=0; i<strlen(geoh); i++)
{
int bits;
for(int j=0; j<32; j++)
if(Base32[j]==geoh[i])
{
bits=j;// 找到第i個字符對應的數
//printf("%d\n",j);
break;
}
for(int j=4; j>=0; j--)
{
int bit = (bits >> j) & 1 ;// 通過位運算取出對應的位
if(odd)
{
double mid = (longitude[0] + longitude[1]) / 2;
longitude[1 - bit] = mid;
}
else
{
double mid = (latitude[0] + latitude[1]) / 2;
latitude[1 - bit] = mid;
}
odd = !odd;
}
}
lat = (latitude[0] + latitude[1]) / 2;
lot = (longitude[0] + longitude[1]) / 2;
}
int main()
{
//freopen("in.in","r",stdin);
int n, m;
scanf("%d%d",&n,&m);
for(int i=1; i<=n; i++)
{
scanf("%lf%lf",&lat,&lot);
//printf("%lf %lf\n",lat,lot);
encode(lat,lot,clegh);
printf("%s\n",geohash);
}
char c=getchar();
memset(geohash,'\0',sizeof(geohash));
for(int i=1; i<=m; i++)
{
scanf("%s",geohash);
c=getchar();
//printf("%s\n",geohash);
docode(geohash);
printf("%lf %lf\n",lat,lot);
}
return 0;
}