[POJ3111]K Best

Posted 2020-10-02 skylee的OI博客

来源：

Northeastern Europe 2005, Northern Subregion

思路：

分数规划经典题。二分分数值$x$，判断其是否满足$\sum (v - w \times m) \geq 0$即可。
针对$v - w \times m$从大到小排序，然后对前$k$项求和，检验是否非负。

 1 #include<cstdio>
 2 #include<cctype>
 3 #include<vector>
 4 #include<algorithm>
 5 #include<functional>
 6 inline int getint() {
 7     char ch;
 8     while(!isdigit(ch=getchar()));
 9     int x=ch^‘0‘;
10     while(isdigit(ch=getchar())) x=(((x<<2)+x)<<1)+(ch^‘0‘);
11     return x;
12 }
13 const double eps=1e-6;
14 const int N=100000;
15 int n,k;
16 struct Jewel {
17     int v,w,id;
18     double sum;
19     bool operator > (const Jewel &x) const {
20         return sum>x.sum;
21     }
22 };
23 Jewel j[N];
24 std::vector<int> ans;
25 inline bool check(const double x) {
26     ans.clear();
27     for(int i=0;i<n;i++) {
28         j[i].sum=j[i].v-j[i].w*x;
29     }
30     std::sort(&j[0],&j[n],std::greater<Jewel>());
31     double sum=0;
32     for(int i=0;i<k;i++) {
33         sum+=j[i].sum;
34         ans.push_back(j[i].id);
35     }
36     return sum>=0;
37 }
38 int main() {
39     n=getint(),k=getint();
40     for(int i=0;i<n;i++) {
41         j[i].v=getint();
42         j[i].w=getint();
43         j[i].id=i+1;
44     }
45     double l=0,r=1000000;
46     while(r-l>=eps) {
47         double mid=(l+r)/2;
48         if(check(mid)) {
49             l=mid;
50         }
51         else {
52             r=mid;
53         }
54     }
55     for(unsigned i=0;i<ans.size();i++) {
56         printf("%d ",ans[i]);
57     }
58     return 0;
59 }