题解

今天下午刚学了模拟退火

借这个题来总结下模拟退火的要注意的问题吧

1 : \(eps\)不要设的太大

2 : 初温\(T\)在2000左右就差不多可以了

3 : 注意题目要求是要求最大值还是最小值,当x<0时\(exp(x)\)的取值范围才是\(0~1\)

4 : 可以在退完火以后再单独从当前最优答案下进行微调

5 : 可以进行多次退火

然后这题就是每次退火就是随机交换序列中的两个数,对序列DP一下就好了

题解

#include<ctime>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
const int M = 25 ;
const int N = 8 ;
const double INF = 1e50 ;
const double EPS = 1e-3 ;
using namespace std ;int n , m ;
int val[M] , e[M] ;
double f[N][M] , p[M] , Sum[M] ;
double bax , Ans = INF ;
inline double Rand() {return (double)((rand() % 101) / 100.0) ;
}
inline double F() {for(int i = 0 ; i <= m ; i ++)for(int j = 0 ; j <= n ; j ++) f[i][j] = INF ;f[0][0] = 0 ;for(int i = 1 ; i <= n ; i ++) Sum[i] = Sum[i - 1] + p[i] ;for(int i = 1 ; i <= m ; i ++)for(int j = i ; j <= n ; j ++)for(int k = i - 1 ; k < j ; k ++)f[i][j] = min(f[i][j] , f[i - 1][k] + (Sum[j] - Sum[k] - bax) * (Sum[j] - Sum[k] - bax)) ;if(f[m][n] < Ans) {Ans = f[m][n] ;for(int i = 1 ; i <= n ; i ++) e[i] = p[i] ;}return f[m][n] ;
}inline void Solve() {for(int i = 1 ; i <= n ; i ++) p[i] = e[i] ;double T = 2000 , W = 0.98 ;double NowAns , PreAns , dlt ;while(T > EPS) {PreAns = F() ;int a = rand() % n + 1 , b = rand() % n + 1 ;while(a == b) b = rand() % n + 1 ;swap(p[a],  p[b]) ;NowAns = F() ; dlt = NowAns - PreAns ;if(exp(-dlt / T) > Rand()) ;else  swap(p[a] , p[b]) ;T *= W ;}for(int i = 1 ; i <= 10000 ; i ++) {int a = rand() % n + 1 , b = rand() % n + 1 ;while(a == b) b = rand() % n + 1 ;swap(p[a] , p[b]) ;F() ;swap(p[a] , p[b]) ;}
}
int main() {srand(time(0)) ;cin >> n >> m ;for(int i = 1 ; i <= n ; i ++) {cin >> val[i] ;bax += val[i] ;p[i] = val[i] ;}bax /= m ; F() ;int Times = 20 ; while(Times--) Solve() ;printf("%.2lf\n",sqrt(Ans / m)) ;return 0 ;
}