POJ1379 Run Away
Posted akmer
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我对模拟退火的理解:https://www.cnblogs.com/AKMer/p/9580982.html
题目传送门:http://poj.org/problem?id=1379
题目意思是要求规定大小平面内某一点,该点到所有给定点的距离中最小距离最大。
类似于费马点做法……不过把统计距离和变成了求距离最小值最大。
费马点不会的可以先看看这篇博客。
BZOJ3680:吊打XXXhttps://www.cnblogs.com/AKMer/p/9588724.html
时间复杂度:(O(能A))
空间复杂度:(O(能A))
爬山算法代码如下:
#include <cmath>
#include <ctime>
#include <cstdio>
#include <algorithm>
using namespace std;
#define sqr(a) ((a)*(a))
int n,lenx,leny;
double ansx,ansy,ans;
int read() {
int x=0,f=1;char ch=getchar();
for(;ch<‘0‘||ch>‘9‘;ch=getchar())if(ch==‘-‘)f=-1;
for(;ch>=‘0‘&&ch<=‘9‘;ch=getchar())x=x*10+ch-‘0‘;
return x*f;
}
struct point {
double x,y;
}p[1001];
double len() {
double x=rand()%200000-100000;
return x/100000;
}
double dis(double x1,double y1,double x2,double y2) {
return sqrt(sqr(x1-x2)+sqr(y1-y2));
}
double calc(double x,double y) {
double tmp=1e18;
for(int i=1;i<=n;i++)
tmp=min(tmp,dis(x,y,p[i].x,p[i].y));//求最小距离最大
if(tmp>ans) {ans=tmp;ansx=x,ansy=y;}
return tmp;
}
void climbhill() {
double now_x=ansx,now_y=ansy;
for(double T=1e7;T>=1e-7;T*=0.998) {
double nxt_x=now_x+len()*T;
double nxt_y=now_y+len()*T;
if(nxt_x<0||nxt_x>lenx||nxt_y<0||nxt_y>leny)continue;
if(calc(now_x,now_y)<calc(nxt_x,nxt_y))
now_x=nxt_x,now_y=nxt_y;
}
}
int main() {
srand(time(0));
int T=read();
while(T--) {
lenx=read(),leny=read(),n=read();
ans=-1e18;ansx=lenx/2;ansy=leny/2;//起始点设在平面中央
for(int i=1;i<=n;i++)
scanf("%lf%lf",&p[i].x,&p[i].y);
climbhill();
printf("The safest point is (%.1lf, %.1lf).
",ansx,ansy);
}
return 0;
}
模拟退火代码如下:
#include <cmath>
#include <ctime>
#include <cstdio>
#include <algorithm>
using namespace std;
#define sqr(x) ((x)*(x))
const double T_0=1e-7,del_T=0.998;
int lenx,leny,n;
double ansx,ansy,ans;
int read() {
int x=0,f=1;char ch=getchar();
for(;ch<‘0‘||ch>‘9‘;ch=getchar())if(ch==‘-‘)f=-1;
for(;ch>=‘0‘&&ch<=‘9‘;ch=getchar())x=x*10+ch-‘0‘;
return x*f;
}
struct point {
double x,y;
}p[1001];
double len() {
double x=rand()%200000-100000;
return x/100000;
}
double dis(double x1,double y1,double x2,double y2) {
return sqrt(sqr(x1-x2)+sqr(y1-y2));
}
double calc(double x,double y) {
double tmp=1e18;
for(int i=1;i<=n;i++)
tmp=min(tmp,dis(x,y,p[i].x,p[i].y));
if(tmp>ans) {ans=tmp;ansx=x;ansy=y;}
return tmp;
}
void Anneal() {
double T=1e7,now_x=ansx,now_y=ansy;
while(T>=T_0) {
double nxt_x=now_x+len()*T;
double nxt_y=now_y+len()*T;
if(nxt_x<0||nxt_x>lenx||nxt_y<0||nxt_y>leny) {
T*=del_T;continue;//不这么写直接卡死循环了……
}
double tmp1=calc(now_x,now_y);
double tmp2=calc(nxt_x,nxt_y);
if(tmp2>tmp1||exp((tmp2-tmp1)/T)*RAND_MAX>rand())
now_x=nxt_x,now_y=nxt_y;
T*=0.998;
}
}
int main() {
srand(time(0));
int T=read();
while(T--) {
lenx=read(),leny=read(),n=read();
ans=-1e18;ansx=lenx/2;ansy=leny/2;
for(int i=1;i<=n;i++)
scanf("%lf%lf",&p[i].x,&p[i].y);
Anneal();
printf("The safest point is (%.1lf, %.1lf).
",ansx,ansy);
}
return 0;
}
程序仅供参考
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