POJ 2976 Dropping tests 01分数规划
来源:互联网 发布:t型截面惯性矩的算法 编辑:程序博客网 时间:2024/05/22 09:09
先从ZP那里粘了点东西过来,再加入自己的理解。
摘:
题目大意就 给定n个二元组(a,b),扔掉k个二元组,使得剩下的a元素之和与b元素之和的比率最大
题目求的是 max(∑a[i] * x[i] / (b[i] * x[i])) 其中a,b都是一一对应的。 x[i]取0,1 并且 ∑x[i] = n - k;
转:那么可以转化一下。 令r = ∑a[i] * x[i] / (b[i] * x[i]) 则必然∑a[i] * x[i] - ∑b[i] * x[i] * r= 0;(条件1)
并且任意的 ∑a[i] * x[i] - ∑b[i] * x[i] * max(r) <= 0 (条件2,只有当∑a[i] * x[i] / (b[i] * x[i]) = max(r) 条件2中等号才成立)
然后就可以枚举r , 对枚举的r, 求Q(r) = ∑a[i] * x[i] - ∑b[i] * x[i] * r 的最大值, 为什么要求最大值呢? 因为我们之前知道了条件2,所以当我们枚举到r为max(r)的值时,显然对于所有的情况Q(r)都会小于等于0,并且Q(r)的最大值一定是0.而我们求最大值的目的就是寻找Q(r)=0的可能性,这样就满足了条件1,最后就是枚举使得Q(r)恰好等于0时就找到了max(r)。而如果能Q(r)>0 说明该r值是偏小的,并且可能存在Q(r)=0,而Q(r)<0的话,很明显是r值偏大的,因为max(r)都是使Q(r)最大值为0,说明不可能存在Q(r)=0了。
自习感受:我用简单的方式在描述一下:a代表a[i]中被选择的数字的和,同理b.这样就会有一个公式a/b = r<==>a-b*r == 0, 这里的r代表的是最大的结果。然后看看此时当r确定时,最优的结果r1与r进行比较,如果a-b*r1==0,说明枚举到此时的r就是最优的解,如果r1>r(a-b*r1 > 0(这里的r1就是枚举出来的结果)),则枚举结果偏大,相反偏小。注意将公式变形可以得到:a-b*r == 0.在这里ai,和bi的选择是同时的,如果a中没有ai,则一定没有bi所以就建立起来联系了啊。这里剔除任意一个数字都是可以的,所以每个数字对应两个状态0剔除,1不剔除。然后我们就设c==a-b*r1,(r1是枚举出来的结果)因为我们想是得r最大,所以我们就在数组c[i] = a[i]-b[i]*r1,中选择最大的n-m个数字,然后就算出一个数字来,根据前面说的进行判断,是否符合要求。
感觉说的有点乱,有点啰嗦了啊,总之就是先建立起来a,b与r的关系。然后比较求出来的r1与要求的r的大小,来二分求r。
Description
In a certain course, you take n tests. If you get ai out of bi questions correct on test i, your cumulative average is defined to be
.
Given your test scores and a positive integer k, determine how high you can make your cumulative average if you are allowed to drop any k of your test scores.
Suppose you take 3 tests with scores of 5/5, 0/1, and 2/6. Without dropping any tests, your cumulative average is . However, if you drop the third test, your cumulative average becomes .
Input
The input test file will contain multiple test cases, each containing exactly three lines. The first line contains two integers, 1 ≤ n ≤ 1000 and 0 ≤ k < n. The second line contains nintegers indicating ai for all i. The third line contains n positive integers indicating bi for all i. It is guaranteed that 0 ≤ ai ≤ bi ≤ 1, 000, 000, 000. The end-of-file is marked by a test case with n = k = 0 and should not be processed.
Output
For each test case, write a single line with the highest cumulative average possible after dropping k of the given test scores. The average should be rounded to the nearest integer.
Sample Input
3 15 0 25 1 64 21 2 7 95 6 7 90 0
Sample Output
83100
#include <algorithm>#include <iostream>#include <stdlib.h>#include <string.h>#include <iomanip>#include <stdio.h>#include <string>#include <queue>#include <cmath>#include <stack>#include <map>#include <set>#define eps 1e-7#define M 1000100//#define LL __int64#define LL long long#define INF 0x3f3f3f3f#define PI 3.1415926535898const int maxn = 1010;using namespace std;int a[maxn], b[maxn];double c[maxn];int main(){ int n, m; while(cin >>n>>m) { if(!n && !m) break; for(int i = 0; i < n; i++) cin >>a[i]; for(int i = 0; i < n; i++) cin >>b[i]; double l = 0.0;//0%; double r = 1.0;//100%; double mid = (l+r)/2; while(abs(r-l) > eps) { mid = (l+r)/2; for(int i = 0; i < n; i++) c[i] = 1.0*a[i]-1.0*b[i]*mid;//算出来所有的ci来 double sum = 0; sort(c, c+n); for(int i = m; i < n; i++) sum += c[i];//求出和来 if(sum < 0)//这说明枚举的r1偏小 r = mid; else l = mid; } mid *= 100; cout<<fixed<<setprecision(0)<<mid<<endl; } return 0;}
- POJ 2976 Dropping tests 01分数规划
- POJ 2976 Dropping tests 01分数规划
- poj-2976-Dropping tests-01分数规划
- POJ 2976 Dropping tests 01分数规划
- 【POJ】2976 Dropping tests 01分数规划
- [poj 2976]Dropping tests 01分数规划
- POJ 2976 Dropping tests (01分数规划)
- POJ 2976 Dropping tests(01分数规划)
- POJ - 2976 Dropping tests 01分数规划
- POJ 2976 Dropping tests 01分数规划
- POJ 2976- Dropping tests -01分数规划
- POJ 2976 Dropping tests 01分数规划
- poj 2976 Dropping tests 01分数规划
- POJ-2976:Dropping tests【01分数规划】
- POJ 2976:Dropping tests 01 分数规划
- POJ 2976 Dropping tests (01分数规划)
- POJ 2976 Dropping tests & 分数规划讲解
- poj Dropping tests 01分数规划
- 日积月累:ScrollView嵌套ListView只显示一行
- hdu 4284 floyd+暴搜
- XtraBackup备份MySQL
- 解决Multiple annotations found at this line: - schema_reference.4: Failed to read schema document '
- 转android gralloc流程分析for msm8960
- POJ 2976 Dropping tests 01分数规划
- glibc改进
- 多功能数字收音机
- 写给静不下心来的朋友们
- POJ 2239 Selecting Courses EK!匈牙利!SAP?
- HBase写数据过程
- CString 如何转化为 char* ?
- 暂停更新博客,请谅解!
- linux0.11调试