[JSOI2004]平衡点

题面在这里

题意

...见链接吧

sol

在此发一篇模拟退火的题解

不得不说luogu的数据真是太良心啦

一句话解释模拟退火:在一个慢慢缩小的范围内随机状态寻找最优解，当转移状态更优时直接接受，当当前状态更优时以一定概率\((exp(dlt/T))\)接受

具体实现请参照代码

#include<bits/stdc++.h>
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<iomanip>
#include<cstring>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define sqr(x) ((x)*(x))
#define pb push_back
#define RG register
#define il inline
using namespace std;
const int mod=1e9+7;
const int N=1010;
typedef unsigned long long ull;
typedef vector<int>VI;
typedef long long ll;
typedef double dd;
il ll read(){
	RG ll data=0,w=1;RG char ch=getchar();
	while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
	if(ch=='-')w=-1,ch=getchar();
	while(ch<='9'&&ch>='0')data=data*10+ch-48,ch=getchar();
	return data*w;
}

int n;dd mn;
struct point{dd x,y,w;}p[N],ans,now;
il dd make(){return rand()%100000/100000.00;}
il dd dis(point a,point b){return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));}
il dd calc(point x){
	RG dd ret=0;
	for(RG int i=1;i<=n;i++)
		ret+=p[i].w*dis(x,p[i]);
	return ret;
}

il void SA(point x){
	RG dd T=1e6;
	while(T>1e-3){//退火过程
		now.x=ans.x+T*(2*make()-1);
		now.y=ans.y+T*(2*make()-1);
		double dlt=calc(ans)-calc(now);
		if(dlt>0||exp(dlt/T)>make())ans=now;
		T*=0.996;
	}
	for(int i=1;i<=50000;++i)//最后在一定范围内搜寻到局部最优解
	{
		now.x=ans.x+T*(2*make()-1);
		now.y=ans.y+T*(2*make()-1);
		if(calc(ans)-calc(now)>0)ans=now;
	}
}

int main()
{
	srand(time(NULL)+rand());
	n=read();
	for(RG int i=1;i<=n;i++)
		scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].w);
	ans=(point){make(),make(),0};mn=ans.w=calc(ans);
	SA(ans);
	printf("%.3lf %.3lf\n",ans.x,ans.y);
	return 0;
}

考试骗分，模拟退火，你值得拥有！