USACO 6.3.3 Cowcycles 暴力

http://train.usaco.org/usacoprob2?a=lQZ5UB8ZNRK&S=cowcycle

题目大意：选出两组齿轮（数量分别不超过5个和10个），设前后轮齿数比为传动比，要求：1、最大传动比至少要是最小传动比的三倍    2、将传动比排序后相邻两者差（difference）的方差要最小

USACO总是出人意料，我原本以为会是一道超难的数学题或者构造题，再不济也得有个很酷的剪枝或是很厉害的规律，结果就是枚举尝试，还不如第一章里的虫洞。。。

/*
ID: frontie1
TASK: cowcycle
LANG: C++
*/
#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

int F, R;
int f_st, f_ed, r_st, r_ed;
int f_gear[15], r_gear[15];
int f_output[20], r_output[20];
double output_var = 0x3f3f3f3f, tem_var;
double arr[100];
double dif[100];
int e = 0;
double total;

void DFS_r(int dep, int st, int ed)
{
    if(dep == R){
        //cout << "judge1 start" << endl;
        if(3*f_gear[0]*r_gear[0] > f_gear[F-1]*r_gear[R-1]) return;

        //cout << "get" << endl;
        e = 0; total = 0;
        for(int i = 0; i < F; ++i){
            for(int k = 0; k < R; ++k){
                arr[e++] = (double)f_gear[i] / r_gear[k];
            }
        }
        sort(arr, arr+e);
        for(int i = 1; i < e; ++i){
            dif[i] = (arr[i]-arr[i-1]);
            total += dif[i];
        }
        total /= (e-1);
        tem_var = 0;
        for(int i = 1; i < e; ++i){
            tem_var += (dif[i] - total) * (dif[i] - total);
        }
        if(tem_var < output_var){
            output_var = tem_var;
            for(int i = 0; i < F; ++i){
                f_output[i] = f_gear[i];
            }
            for(int i = 0; i < R; ++i){
                r_output[i] = r_gear[i];
            }
        }
        return;
    }
    else if(st > ed) return;
    else{
        for(int i = st; i <= ed; ++i){
            r_gear[dep] = i;
            DFS_r(dep+1, i+1, ed);
        }
    }
}

void DFS_f(int dep, int st, int ed)
{
    if(dep == F){
        //cout << "R start" << endl;
        DFS_r(0, r_st, r_ed);
        return;
    }
    else if(st > ed) return;
    for(int i = st; i <= ed; ++i){
        f_gear[dep] = i;
        DFS_f(dep+1, i+1, ed);
    }
}

int main()
{
    freopen("cowcycle.in", "r", stdin);
    freopen("cowcycle.out", "w", stdout);

    cin >> F >> R;
    cin >> f_st >> f_ed >> r_st >> r_ed;

    DFS_f(0, f_st, f_ed);

    for(int i = 0; i < F; ++i){
        cout << f_output[i];
        if(i != F-1) cout << ' ';
        else cout << endl;
    }

    for(int i = 0; i < R; ++i){
        cout << r_output[i];
        if(i != R-1) cout << ' ';
        else cout << endl;
    }

    return 0;
}

悄悄去看了USACO官方题解，原来还能哈希提速的，但是不想细看了，没有需求就没有动力（笑）