题意:
n个物品,每个物品有v(i)和w(i),要求选出k个物品,问sigma(v)/sigma(w)的最大值是多少
思路:
二分解法:
设答案为x,则x=sigma(v)/sigma(w)
如果要满足式子,那么x<=sigma(v)/sigma(w)
把右边的w乘到左边,则式子变为:
x*sigma(w)<=sigma(v)
移项得:
sigma(v)-x*sigma(w)>=0
二分x,check方法为令d(i)=v(i)-x*w(i),排序一下贪心第选择前k大得,判断和是否大于0。
牛顿迭代:
待更
code1:
二分
ps:g++提交ac,c++提交超时
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxm=1e5+5;
const double eps=1e-6;
struct Node{
int id;
double v;
}e[maxm];
int v[maxm],w[maxm];
int n,k;
bool cmp(Node a,Node b){
return a.v>b.v;
}
bool check(double mid){
for(int i=1;i<=n;i++){
e[i].id=i;
e[i].v=v[i]-mid*w[i];
}
sort(e+1,e+1+n,cmp);
double sum=0;
for(int i=1;i<=k;i++){
sum+=e[i].v;
}
return sum>eps;
}
signed main(){
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++){
scanf("%d%d",&v[i],&w[i]);
}
double l=0,r=1e6;
while(r-l>eps){
double mid=(l+r)/2;
if(check(mid)){
l=mid;
}else{
r=mid;
}
}
for(int i=1;i<=k;i++){
printf("%d",e[i].id);
if(i!=k)printf(" ");
else printf("\n");
}
return 0;
}