USACO Section 4.2 Cowcycles - 枚举~

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开始想复杂了...纠结了很久..后来发现数据范围很小阿...枚举所有情况也就那么多...所以就两层递归暴力枚举..结果就过了....话说这题主要还是题意要理解清楚...这里的mean和variance中的Xi..是指排好序的F[i]/R[i]的前后相差的数...其实样例下面的说明也体现了...

Program:

/*  ID: zzyzzy12   LANG: C++   TASK: cowcycle*/      #include<iostream>      #include<istream>  #include<stdio.h>      #include<string.h>      #include<math.h>      #include<stack>#include<map>#include<algorithm>      #include<queue>   #define oo 1000000000  #define ll long long  #define pi (atan(2)+atan(0.5))*2 using namespace std; int F,R,F1,F2,R1,R2,f[6],r[11];int ansF[6],ansR[11];double MinData,mean,x[1001],t;void judge(){      int i,j,num=0;       for (i=1;i<=F;i++)         for (j=1;j<=R;j++)            x[++num]=f[i]*1.0/r[j];       sort(x+1,x+1+num);      num--;      for (i=1;i<=num;i++) x[i]=x[i+1]-x[i];       mean=0;      for (i=1;i<=num;i++) mean+=x[i];      mean/=num; t=0;      for (i=1;i<=num;i++) t+=(x[i]-mean)*(x[i]-mean);      t/=num;            if (t<MinData)       {             MinData=t;             for (i=1;i<=F;i++) ansF[i]=f[i];             for (i=1;i<=R;i++) ansR[i]=r[i];              }      return; }void DFS2(int p,int h){      if (p>R)      {            if (f[F]*r[R]<3*f[1]*r[1]) return;            judge();            return;              }      for (int i=h;i<=R2;i++)      {            r[p]=i;            DFS2(p+1,i+1);      }}void DFS1(int p,int h){      if (p>F)       {            DFS2(1,R1);            return;      }      for (int i=h;i<=F2;i++)      {            f[p]=i;            DFS1(p+1,i+1);          }}int main()  {        freopen("cowcycle.in","r",stdin);         freopen("cowcycle.out","w",stdout);        scanf("%d%d%d%d%d%d",&F,&R,&F1,&F2,&R1,&R2);      int i,j;        MinData=oo;      DFS1(1,F1);      printf("%d",ansF[1]);      for (i=2;i<=F;i++) printf(" %d",ansF[i]);      printf("\n%d",ansR[1]);      for (i=2;i<=R;i++) printf(" %d",ansR[i]);      printf("\n");      return 0;     }