【USACO4.2.4】奶牛自行车

其实是个水题，就按照题目意思来。

但是我题目意思看错了3次…… 写了3个多小时……

Executing...   Test 1: TEST OK [0.008 secs, 3396 KB]   Test 2: TEST OK [0.095 secs, 3396 KB]   Test 3: TEST OK [0.054 secs, 3396 KB]   Test 4: TEST OK [0.127 secs, 3396 KB]   Test 5: TEST OK [0.267 secs, 3396 KB]   Test 6: TEST OK [0.165 secs, 3396 KB]   Test 7: TEST OK [0.292 secs, 3396 KB]   Test 8: TEST OK [0.475 secs, 3396 KB]All tests OK.

for (back_min = r1; back_min + R - 1 <= r2; ++ back_min)// 穷举后轮最小值back_min{for (back_max = back_min + R - 1; back_max <= r2; ++ back_max) //后轮最大值back_max{for (front_min = f1; front_min + F - 1 <= f2; ++ front_min) //前轮最小值，{for (front_max = front_min + F - 1; front_max <= f2; ++ front_max) //前轮最大值{if (front_min * back_min * 3 > front_max * back_max)continue; //直接不符合要求，退出usd1[1] = front_min;usd2[1] = back_min;usd2[R] = back_max;usd1[F] = front_max;if (F == 1) dfs_front(front_min, 0);else dfs_front(front_min, 1);}}}}

强行穷举前后轮最大值最小值范围，然后根据题目的条件的，大小轮子比率3倍，来直接剪掉一些范围，然后在范围内进行枚举，求解

我是用2个DFS强行穷举出所有情况，然后计算方差。

先看成是选择轮子的方差了……然后又看成是轮子比率的方差，

结果求的是—— 轮子比率排序后的差的方差……

/*TASK:cowcycleLANG:C++*/#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;//F,R为1的情况暂时没有考虑int F, R;int f1, f2, r1, r2;void init(){cin >> F >> R;cin >> f1 >> f2;cin >> r1 >> r2;}int back_min, back_max, front_min, front_max;//int ansint usd1[100], usd2[100], tot1;int output_usd1[100], output_usd2[100];double ans=10000000000;double p[1000],pp[1000];int tp, tt;inline void check(){tp = tt = 0;for (int i = 1; i <= F; ++ i)for (int j = 1; j <= R; ++ j)pp[++tp] = (double)usd1[i] / (double)usd2[j];sort(pp + 1, pp + 1 + tp);for (int i = 1; i < tp; ++ i)p[i] = pp[i + 1] - pp[i];--tp;double ave,tot(0), tmp(0);for (int i = 1; i <= tp; ++ i)tot += p[i];ave = tot / tp;for (int i = 1; i <= tp; ++ i)tmp += (p[i] - ave) * (p[i] - ave);tmp /= tp;if (tmp == ans){for (int i = 1; i <= F; ++ i){if (usd1[i] < output_usd1[i])goto ok1;if (usd1[i] > output_usd1[i])return;}for (int i = 1; i <= R; ++ i){if (usd2[i] < output_usd2[i])goto ok1;if (usd2[i] > output_usd2[i])return;}}if (tmp < ans){ans = tmp;ok1:;memmove(output_usd1 + 1, usd1 + 1, sizeof(int) * F);memmove(output_usd2 + 1, usd2 + 1, sizeof(int) * R);}}void dfs_back(int k,int tot){if (back_max - k + 1 < R - tot)return;if (tot == R - 1){check();return;}usd2[tot + 1] = k;dfs_back(k + 1, tot + 1);dfs_back(k + 1, tot);}void dfs_front(int k, int tot){if (front_max - k + 1 < F - tot)return; //不够填充的，returnif (tot == F - 1) //最后一格直接填进去{dfs_back(back_min + 1, 1);return ;}usd1[tot + 1] = k;//第tot个数字选择kdfs_front(k + 1, tot + 1);dfs_front(k + 1, tot);}void doit(){for (back_min = r1; back_min + R - 1 <= r2; ++ back_min)// 穷举后轮最小值back_min{for (back_max = back_min + R - 1; back_max <= r2; ++ back_max) //后轮最大值back_max{for (front_min = f1; front_min + F - 1 <= f2; ++ front_min) //前轮最小值，{for (front_max = front_min + F - 1; front_max <= f2; ++ front_max) //前轮最大值{if (front_min * back_min * 3 > front_max * back_max)continue; //直接不符合要求，退出usd1[1] = front_min;usd2[1] = back_min;usd2[R] = back_max;usd1[F] = front_max;if (F == 1) dfs_front(front_min, 0);else dfs_front(front_min, 1);}}}}for (int i = 1; i < F; ++ i)cout<<output_usd1[i]<<" ";cout<<output_usd1[F]<<endl;for (int i = 1; i < R; ++ i)cout<<output_usd2[i]<<" ";cout<<output_usd2[R]<<endl;}int main(){freopen("cowcycle.in", "r",stdin);freopen("cowcycle.out","w",stdout);init();doit();return 0;}