[POJ2976] Dropping tests

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题意:给出长度相等的数组a和b,定义他们的和为$dfrac{a_1+a_2+...+a_n}{b_1+b_2+...+b_n}$。现在可以舍弃k对元素(一对即$a[i]和b[i]$),问最大的和是多少?

解题思路

01分数规划入门题(并没有学过,看到hy大佬在刷因此也去学了下)

问题可以转化为数组中的每个元素选或不选,也就可以认为每一个元素都乘上一个$x[i], x[i] ∈ {0, 1}$

因此问题可以转化为$ans = dfrac{sumlimits_{i = 1}^{n}a[i] * x[i]}{sumlimits_{i = 1}^{n}b[i] * x[i]}$

将除法转化为加法$sumlimits_{i = 1}^{n}a[i] * x[i] - ans * sumlimits_{i = 1}^{n}b[i] * x[i] = 0$

合并得$sumlimits_{i = 1}^{n}(a[i]-ans*b[i])*x[i] = 0$

当$x$数组的取值确定时,可以发现函数$f(r) = sumlimits_{i = 1}^{n}(a[i]-r*b[i])*x[i]$是减函数,因此可以二分$r$。当前取到的$r$能够满足$sumlimits_{i = 1}^{n}(a[i]-r*b[i])*x[i] geq 0$即为可行,为了满足此条件,肯定要让选择的那些元素的和越大越好,因此可以建立数组$d[i] = a[i]-r*b[i]$并排序,选择最大的加起来验证是否大于等于0.

Code

long long

/*By DennyQi 2018.8.12*/
#include <cstdio>
#include <queue>
#include <cstring>
#include <algorithm>
#define  r  read()
#define  Max(a,b)  (((a)>(b)) ? (a) : (b))
#define  Min(a,b)  (((a)<(b)) ? (a) : (b))
using namespace std;
typedef long long ll;
#define int long long
const int MAXN = 10010;
const int MAXM = 27010;
const int INF = 1061109567;
inline int read(){
    int x = 0; int w = 1; register int c = getchar();
    while(c ^ - && (c < 0 || c > 9)) c = getchar();
    if(c == -) w = -1, c = getchar();
    while(c >= 0 && c <= 9) x = (x << 3) + (x << 1) + c - 0, c = getchar(); return x * w;
}
struct Score{
    int idx; double sc;
}s[MAXN];
int N,K;
int a[MAXN],b[MAXN];
double L,R,Mid,d[MAXN];
inline bool comp(const Score& a, const Score& b){
    return a.sc < b.sc;
}
inline bool judge(double _r){
    for(int i = 1; i <= N; ++i){
        d[i] = (double)((double)(a[i]) - (double)(1.0*_r*b[i]));
    }
    sort(d+1,d+N+1);
    double res = 0.0;
    for(int i = N; i > K; --i){
        res += d[i];
    }
    return res >= 0.0;
}
#undef int
int main(){
#define int long long
    for(;;){
        N = r, K = r;
        if(!N && !K) break;
        for(int i = 1; i <= N; ++i) a[i] = r;
        for(int i = 1; i <= N; ++i) b[i] = r;
        L = 0.000, R = 9999999999.999;
        while(R - L >= 1e-8){
            Mid = (L + R) / 2.000;
            if(judge(Mid)){
                L = Mid;
            }
            else{
                R = Mid;
            }
        }
        for(int i = 1; i <= N; ++i){
            s[i] = (Score){i, (double)((double)(a[i]) - (double)(1.0*L*b[i]))};
        }
        sort(s+1,s+N+1,comp);
        int fz=0,fm=0;
        for(int i = N; i > K; --i){
            fz += a[s[i].idx];
            fm += b[s[i].idx];
        }
        double rs = (double)fz/(double)fm;
        printf("%.0f
", rs * 100);
    }
    return 0;
}

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