原题地址
比平衡点还裸的模拟退火模板题,但是窝就是不会打,
太菜了
n个数分m组,让方差最小,就是对于ai把ai放到最小的xi中(1<=x<=n)
对于n个数随机重组
退火很迷的我对着大佬的参数一个一个试下来(真开心嘤嘤嘤)
走好
// luogu-judger-enable-o2
#include <bits/stdc++.h>
using namespace std;
int n,m,a[25],grp[25],sum[25];
double ans,avg;
const double d = 0.992;
double get_val()
{
double val = 0;
for (int i = 1;i <= m;i++)
val += (sum[i] - avg) * (sum[i] - avg);
return val / m;
}
void ch_grp(int pos,int target)
{
sum[grp[pos]] -= a[pos];
sum[target] += a[pos];
grp[pos] = target;
}
void SA()
{
double T = 3000,cur_ans = ans;
while (T > 1e-10)
{
int pos = rand() % n + 1,target = rand() % m + 1;
int origin = grp[pos];
ch_grp(pos,target);
double nans = get_val(),diff = nans - cur_ans;
if (diff < 0)
{
cur_ans = nans;if (cur_ans < ans) ans = cur_ans;
}
else if (RAND_MAX * exp(-diff / T) <= rand()) ch_grp(pos,origin);
T *= d;
}
}
int main ()
{
srand(time(0));
scanf("%d%d",&n,&m);
for (int i = 1;i <= n;i++)
{
scanf("%d",&a[i]);avg += a[i];
}
avg /= m;
for (int i = 1;i <= n;i++)
{
grp[i] = rand() % m + 1;sum[grp[i]] += a[i];
}
ans = get_val();
for (int i = 1;i <= 1000;i++) SA();
printf("%.2lf",sqrt(ans));
return 0;
}
放弃吧,假的参数,我才不会给你真的呢。