poj 2976: Dropping tests（01分数规划--Dinkelbach算法）

Dropping tests

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 13323 Accepted: 4675

Description

In a certain course, you take n tests. If you get a_i out of b_i 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 n integers indicating a_i for all i. The third line contains n positive integers indicating b_i for all i. It is guaranteed that 0 ≤ a_i ≤ b_i ≤ 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

题意：

给你n组数，每组数中有两个数a[i]和b[i]，让你在这n组数中选出n-k组使得100*∑a[i]/∑b[i]值最大

经典01分数规划问题：

http://blog.csdn.net/jaihk662/article/details/77505318

还是这个公式，x[i]表示选还是不选

先给出二分思路：

二分cnt，然后求出所有的d[i]，按d[i]从大到小排序选择前n-k个并求和，如果大于0说明可能不是最优解

再说Dinkelbach算法：

先随机一个cnt，然后求出所有的d[i]，按d[i]从大到小排序选择前n-k个求出∑a[i]/∑b[i]，如果∑a[i]/∑b[i]==cnt，那么说明是最优解，否则令cnt=∑a[i]/∑b[i]继续，直到求出最优解（一般优于二分）

其实这题还有一个贪心思路：算出所有的d[i] = a[i]/b[i]，也就是比率，然后按这个从大到小排序，选最大的k个

但这是错的，一个很简单的反例：

3 1

100 1 100

100 5 200

正确答案应该是(101/105)*100 ≈ 96

#include<stdio.h>#include<algorithm>#include<math.h>using namespace std;typedef struct Res{double a, b;double d;bool operator < (const Res &b) const{if(d>b.d)return 1;return 0;}}Res;Res s[1005];int main(void){int i, n, k;double cnt, sa, sb;while(scanf("%d%d", &n, &k), n!=0 || k!=0){k = n-k;for(i=1;i<=n;i++)scanf("%lf", &s[i].a);for(i=1;i<=n;i++)scanf("%lf", &s[i].b);cnt = 0;while(1){for(i=1;i<=n;i++)s[i].d = s[i].a-cnt*s[i].b;sort(s+1, s+n+1);sa = sb = 0;for(i=1;i<=k;i++){sa += s[i].a;sb += s[i].b;}if(fabs(sa/sb-cnt)<0.00001)break;cnt = sa/sb;}printf("%.0f\n", cnt*100);}return 0;}