模拟退火板子

Posted 2021-02-20 tztqwq

参数

初始温度 (tmp)， 降温系数 (Delta)。

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接受不优解的条件

[e^{frac{-当前解与最优解的差}{tmp}} * rand\_max < rand() ]

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卡时间代码

while((double)clock()/CLOCKS_PER_SEC < 时限) 模拟退火();

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板子

double ans=1e18;
struct ANS_type {
	/*
		balabala
	*/
} ansnode;

double t;
const double delta = 0.993;
void SA() {
	t = 2000.0;
	ANS_type nownode = ansnode;
	//可能要SA多次， 从上一次结束的地方继续
	while(t>1e-14) {
		/*
			随机生成一个新解， 当前解与新解的距离 通常与温度有关 
		*/
		if(calc_ans(新解) 优于 ans) {
			ans = calc_ans(新解);
			nownode = ansnode = 新解 
		} else if(exp(-当前解与最优解的差/t)*RAND_MAX > rand()) {
			nownode = 新解; 
		}
		t *= delta;
	}
}

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例题代码

#[JSOI2004]平衡点 / 吊打XXX

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1005;
int n;
int x[maxn], y[maxn], w[maxn];

double calcans(double X, double Y) {
	double res = 0;
	for(int i=1;i<=n;++i) {
		double dx = x[i]-X, dy=y[i]-Y;
		res += sqrt(dx*dx+dy*dy) * w[i];
	}
	return res;
}

double ans=1e18, ansx, ansy;
double t;
const double delta = 0.994;
void SA() {
	double X=ansx, Y=ansy;
	t = 3000;
	while(t>1e-14) {
		double nowx = X + (rand()*2-RAND_MAX)*t;
		double nowy = Y + (rand()*2-RAND_MAX)*t;
		double nowans = calcans(nowx, nowy);
		double Delta = nowans - ans;
		if(Delta < 0) {
			ansx = X = nowx, ansy = Y = nowy;
			ans = nowans;
		} else if(exp(-Delta/t)*RAND_MAX > rand()) X=nowx, Y=nowy;
		t*=delta;
	}
}

int main()
{
	srand(13333337); srand(rand()); srand(rand());
	
	int sx=0, sy=0;
	scanf("%d", &n);
	for(int i=1;i<=n;++i)
		scanf("%d%d%d", &x[i],&y[i],&w[i]), sx+=x[i], sy+=y[i];
	ansx = (double)sx/n, ansy=(double)sy/n;
	SA();
	SA();
	SA();
	printf("%.3lf %.3lf
", ansx, ansy);
	return 0;
}