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;
}