POJ2976:Dropping tests(01分数规划入门)
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 n integers 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 1 5 0 2 5 1 6 4 2 1 2 7 9 5 6 7 9 0 0
Sample Output
83 100
Hint
To avoid ambiguities due to rounding errors, the judge tests have been constructed so that all answers are at least 0.001 away from a decision boundary (i.e., you can assume that the average is never 83.4997).
题意:有几门成绩,现在要求选N-K门,使得平均成绩最高,求最高平均成绩。
思路:01分数规划。
有三种需要掌握的01分数规划,之前最大流的时候遇到过,现在有些想法,有空再来实现。
#include<cstdio> #include<cstdlib> #include<iostream> #include<cstring> #include<algorithm> using namespace std; const double eps=1e-6; int N,K; struct in { double x,y; double res; bool friend operator <(in a,in b){ return a.res>b.res; } }a[1010]; bool check(double Mid) { for(int i=1;i<=N;i++) a[i].res=a[i].x-Mid*a[i].y; sort(a+1,a+N+1); double tx=0; for(int i=1;i<=K;i++) tx+=a[i].res; if(tx>=-eps) return true; return false; } int main() { while(~scanf("%d%d",&N,&K)&&(N||K)){ K=N-K; for(int i=1;i<=N;i++) scanf("%lf",&a[i].x),a[i].x*=100.0; for(int i=1;i<=N;i++) scanf("%lf",&a[i].y); double L=0,R=100,Mid,ans=0; while(R-L>eps){ Mid=(L+R)/2; if(check(Mid)) ans=Mid,L=Mid+eps; else R=Mid-eps; } printf("%.0lf\n",ans); } return 0; }