模拟退火求解TSP问题

模拟退火算法步骤

1.寻找下一个解
2.计算下一个解的能量
3.决定是否接受这个解
4.降温

算法模板

double randfloat() {
	return rand()/(RAND_MAX+0.0);
}

double T0 = 1000000,Tk = 1,T = T0,d = 0.9999;
int x = initx();//当前解(初始解)
int ansE,nowE;//全局最优解的能量,当前解的能量

ansE = nowE = E(x);

while(T > Tk) {
	//1.产生新解
	int newx = generate(x);
	//2.计算新解的能量
	int newE = E(newx);
	//3.确定是否接受
	if(newE < nowE || randfloat() > exp((nowE-newE)/T)){
		nowE = newE;
		x = newx;
	}
	//4.降温
	T *= d;
	//5.更新全局最优解
	Optimize(ansE,newE);
}

TSP问题代码

交题链接
http://acm.zjnu.edu.cn/CLanguage/showproblem?problem_id=1420

#include <iostream>
#include <algorithm>
#include <cstring>
#include <ctime>
#include <cmath>
#include <assert.h>
#define pr(x) std::cout << #x << ':' << x << std::endl
#define rep(i,a,b) for(int i = a;i <= b;++i)
#define int long long
int G[21][21];
int n,m;
int run(int x[]){
	int ans = 0;
	rep(i,1,n) {
		ans += G[x[i-1]][x[i%n]];
	}
	return ans;
}

double randfloat() {
	return rand()/(RAND_MAX+0.0);
}

int solve() {
	srand(time(NULL));
	int x[22];
	rep(i,0,n-1) x[i] = i+1;
	double T0 = 100000000,Tk = 1,T = T0,d = 0.99;
	int ans = run(x);
	int preans = ans;
	while(T > Tk) {
		//1.产生新解
		int a = rand()%n;
		int b = rand()%n;
		//std::swap(x[a],x[b]);
		std::reverse(x+a,x+b+1);
		//2.计算下一个解的能量
		int newans = run(x);
		Min(ans,newans);
		//3.确定是否接受
		if(newans >= preans) {
			if(randfloat() > exp((preans-newans)/T))
				//std::swap(x[a],x[b]);
				std::reverse(x+a,x+b+1);
			else preans = newans;
		}
		else preans = newans;
		//4.降温
		T *= d;
	}
	return ans;
}

signed main() {
	std::cin >> n >> m;
	//setinf(G);
	if(n == 1) return 0*puts("0");
	rep(i,1,n) rep(j,1,n) {
		G[i][j] = 1e11;
	}
	rep(i,1,m) {
		int a,b,c;
		std::cin >> a >> b >> c;
		Min(G[a][b],c);
		Min(G[b][a],c);
	}
	std::cout << solve() << std::endl;
}