BZOJ 2428 [HAOI2006]均分数据 模拟退火

题目大意：已知N个正整数：A1、A2、……、An 。今要将它们分成M组，使得各组数据的数值和最平均，即各组的均方差最小。

模拟退火乱搞…
这个参数真的是玄学…

#include <cstdio>#include <cmath>#include <algorithm>#define N 25#define INF 1e20using namespace std;inline int RAND() { return /*(rand()<<15)+*/rand(); }inline double sq(double x) { return x*x; }int n,m,b[N];double a[N],sum[N],ave,ans=INF;double SAA() { //Simulate Anneal Arithmetic    for(int i=1;i<=m;++i) sum[i]=0;    for(int i=1;i<=n;++i)        b[i]=RAND()%m+1, sum[b[i]]+=a[i];    double T=1e5,last,now=0;    for(int i=1;i<=m;++i) now+=sq(sum[i]-ave);    for(;T>1e-5;T*=0.75) {        last=now;        int x=RAND()%n+1,y=0;        if(T<=1e4) y=rand()%m+1;        for(int i=1;i<=m;++i)            if(sum[i]<sum[y]) y=i;        now+=a[x]*(a[x]+sum[y]-sum[b[x]])*2;        if(now>last && T<RAND()%100000) now=last;        else sum[b[x]]-=a[x], sum[y]+=a[x], b[x]=y;    }    return now;}int main() {    srand(20000301);    scanf("%d%d",&n,&m);    for(int i=1;i<=n;++i) scanf("%lf",a+i), ave+=a[i];    sum[0]=INF, ave/=m;    for(int i=1;i<=50000;++i) ans=min(ans,SAA());    printf("%.2f\n",sqrt(ans/m));    return 0;}