模拟退火学习

巨佬的讲解
P1337 [JSOI2004]平衡点 / 吊打XXX
题意：
桌子上有 $n$ 个孔，每个孔都有一个一根绳穿过，下边吊着一个重物，重物坐标 $x,y$，这些绳打结在一起，绳子是完全弹性的，孔是足够小的，保证绳结最多是卡在某个洞口处。求最后绳结的位置
保留三位小数。
思路：
模拟退火
code：

#include<bits/stdc++.h>
#define endl '\n'
#define ll long long
#define ull unsigned long long
#define ld long double
#define all(x) x.begin(), x.end()
#define eps 1e-15
using namespace std;
const int maxn = 2e5 + 9;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const double down = 0.996;//徐徐降温 
const double MAX_TIME = 0.9;

ll n, m;
struct node{
	double x, y, w;
}a[maxn];
double ansx, ansy, answ;

double energy(double x, double y)//根据物理学知识,能量总和越小越稳定
{
	double ans = 0, dx, dy;
	for(int i = 1; i <= n; ++i)
	{
		dx = x - a[i].x;
		dy = y - a[i].y;
		ans += sqrt(dx * dx + dy * dy) * a[i].w;
	}
	return ans;
}
void sa()
{
	double t = 3000;//温度要足够高
	while(t > eps)
	{
		double ex = ansx + (rand() * 2 - RAND_MAX) * t;// 每次随机的数跟t建立联系
		double ey = ansy + (rand() * 2 - RAND_MAX) * t;
		double ew = energy(ex, ey);
		double diff = ew - answ;
		if(diff < 0) ansx = ex, ansy = ey, answ = ew;// 更新答案 
		else if(exp(-diff / t) * RAND_MAX > rand())// 根据多项式概率更新答案 
			ansx = ex, ansy = ey;//  实测不进行这个更新也能AC 
		t *= down;
	}
}
void work()
{
	cin >> n;
	for(int i = 1; i <= n; ++i){
		cin >> a[i].x >> a[i].y >> a[i].w;
		ansx += a[i].x; ansy += a[i].y;
	}
	ansx /= n;  ansy /= n;//以平均数作为初始答案
	answ = energy(ansx, ansy);
	while ((double)clock() / CLOCKS_PER_SEC * 1.0 < MAX_TIME)// MAX_TIME是执行时间，根据需要进行调整
        sa();
	cout << fixed << setprecision(3) << ansx << " " << ansy;
}

int main()
{
//	srand(time(0));
	ios::sync_with_stdio(0);
//	int TT;cin>>TT;while(TT--)
	work();
	return 0;
}

P2216 [HAOI2007]理想的正方形
题意：
在 $a\times b$ 的矩阵中找到一个 $n\times n$ 的矩阵，使得矩阵内最大值与最小值差值最大
思路：
巨佬的讲解
这里只谈模拟退火，这种设计整数随机化搜索的不是很容易 $ac$ ，毕竟是假做法
按照这位巨佬的搜索，能拿 $60-80$ 分左右，开 $O^2$ 更容易拿高分
正解是单调队列
code：

#include<bits/stdc++.h>
#define endl '\n'
#define ll long long
#define ull unsigned long long
#define ld long double
#define all(x) x.begin(), x.end()
#define eps 1e-8
using namespace std;
const int maxn = 1e3 + 9;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const double down = 0.996;
const double MAX_TIME = 0.9;
ll n, m, a, b;
int ansx, ansy, ans = inf;
int w[maxn][maxn];

int cal(int x, int y)
{
	int Max = -inf, Min = inf;
	for(int i = x - n + 1; i <= x; ++i)
		for(int j = y - n + 1; j <= y; ++j)
			Max = max(Max, w[i][j]),
			Min = min(Min, w[i][j]);
	return Max - Min;
}
void sa()
{
	double t = max(a, b);
	int x = ansx, y = ansy;
	while(t > 0.1)
	{
		int dx = ((x + (rand() * 2 - RAND_MAX)) % (int)round(t * 10)) % a + 1;
		int dy = ((y + (rand() * 2 - RAND_MAX)) % (int)round(t * 10)) % b + 1;
		t *= down;
		if(dx < n || dy < n || dx > a || dy > b) continue;
		double now = cal(dx, dy);
		double diff = now - ans;
		if(diff < 0) x = dx,  y = dy, ansx = dx, ansy = dy, ans = now;
		else if(exp(-diff / t) * RAND_MAX > rand())
			x = dx, y = dy;
	}
}
void work()
{
	srand(time(NULL));
	cin >> a >> b >> n;
	for(int i = 1; i <= a; ++i)
		for(int j = 1; j <= b; ++j)
			cin >> w[i][j];
	ansx = ansy = n;
	while ((double)clock() / CLOCKS_PER_SEC * 1.0 < MAX_TIME) sa();
	cout << ans << endl;
}

int main()
{
	ios::sync_with_stdio(0);
//	cout << rand() << " " << RAND_MAX << endl;
//	int TT;cin>>TT;while(TT--)
	work();
	return 0;
}