POJ 2420 A Star not a Tree?(二维费马点)
题目:就是求多边形的费马点,输出最小的距离。
http://poj.org/problem?id=2420
做法:随机化变步长贪心法(模拟退火???)
首先随机选出一点,我直接取了0,0
然后选定一个步长,往4个方向开始找,如果更优则继续,否则降低步长,直到满足题目要求精度
也可以8个方向等等
[cpp]
#include<iostream>
#include<fstream>
#include<iomanip>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<vector>
#include<sstream>
#include<cassert>
#define LL long long
#define eps 1e-6
#define inf 1ll<<50
#define zero(a) fabs(a)<eps
#define N 20005
using namespace std;
struct Point{
double x,y;
}p[105],cur,pre;
int n;
int way[4][2]={0,1,0,-1,1,0,-1,0};
double dist(Point p1,Point p2){
return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
double get_dist(Point cen){
double sum=0;
for(int i=0;i<n;i++)
sum+=dist(cen,p[i]);
return sum;
}
int main(){
while(scanf("%d",&n)!=EOF){
for(int i=0;i<n;i++)
scanf("%lf%lf",&p[i].x,&p[i].y);
pre.x=0;pre.y=0;
double step=100,best=get_dist(pre);
while(step>0.2){
bool ok=true;
while(ok){
ok=false;
for(int i=0;i<4;i++){
cur.x=way[i][0]*step+pre.x;
cur.y=way[i][1]*step+pre.y;
double dis=get_dist(cur);
if(dis<best){best=dis;pre=cur;ok=true;}
}
}
step/=2.0;
}
printf("%d\n",(int)(best+0.5)*100/100);
}
return 0;
}
补充:软件开发 , C++ ,