POJ 3111

K Best
Time Limit: 8000MS   Memory Limit: 65536K
Total Submissions: 5177   Accepted: 1411
Case Time Limit: 2000MS   Special Judge

Description

Demy has n jewels. Each of her jewels has some value vi and weight wi.

Since her husband John got broke after recent financial crises, Demy has decided to sell some jewels. She has decided that she would keep k best jewels for herself. She decided to keep such jewels that their specific value is as large as possible. That is, denote the specific value of some set of jewels S = {i1, i2, …, ik} as

.

Demy would like to select such k jewels that their specific value is maximal possible. Help her to do so.

Input

The first line of the input file contains n — the number of jewels Demy got, and k — the number of jewels she would like to keep (1 ≤ kn ≤ 100 000).

The following n lines contain two integer numbers each — vi and wi (0 ≤ vi ≤ 106, 1 ≤ wi ≤ 106, both the sum of all vi and the sum of all wi do not exceed 107).

Output

Output k numbers — the numbers of jewels Demy must keep. If there are several solutions, output any one.

Sample Input

3 2
1 1
1 2
1 3

Sample Output

1 2

Source

Northeastern Europe 2005, Northern Subregion
 
二分平均值
x为平均值,判断的时候 计算每个数的 (v[i] - w[i] * x) ,排序,找出最大的k个数,加起来,如果大于等于0 则往上找,否则往下
 
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <utility>
 6 
 7 using namespace std;
 8 
 9 #define maxn 100005
10 #define eps 1e-8
11 typedef pair<double,int> pii;
12 
13 int w[maxn],v[maxn];
14 double f[maxn];
15 pii ans[maxn];
16 int n,k;
17 
18 
19 bool judge(double x) {
20         for(int i = 1; i <= n; ++i) {
21                 f[i] = (v[i] - x * w[i]);
22         }
23         sort(f + 1,f + n + 1);
24 
25         double sum = 0;
26         for(int i = n; i >= n - k + 1; --i) {
27                 sum += f[i];
28         }
29 
30         return sum >= 0;
31 
32 }
33 
34 void output(double x) {
35 
36         for(int i = 1; i <= n; ++i) {
37                 ans[i] = make_pair(v[i] - x * w[i],i);
38         }
39         sort(ans + 1,ans + n + 1);
40 
41         printf("%d",ans[n].second);
42         for(int i = n - 1; i >= n - k + 1; --i) {
43                 printf(" %d",ans[i].second);
44         }
45 
46         printf("\n");
47 
48 
49 }
50 
51 void solve() {
52         double l = 0,r = 0;
53         for(int i = 1; i <= n; ++i) {
54                 r += v[i];
55         }
56 
57         int num = 0;
58         while(num++ < 50 ) {
59                 //printf(" l = %f r = %f\n",l,r);
60                 //num++;
61                 double mid = (l + r) / 2;
62                 if(judge(mid)) {
63                         l = mid;
64                 } else {
65                         r = mid;
66                 }
67 
68         }
69 
70         //printf("%f\n",l);
71         output(l);
72 }
73 int main()
74 {
75 
76     //freopen("sw.in","r",stdin);
77 
78     while(~scanf("%d%d",&n,&k))
79     {
80         for(int i = 1; i <= n; ++i) {
81             scanf("%d%d",&v[i],&w[i]);
82         }
83 
84         solve();
85     }
86 
87     return 0;
88 }
View Code

 

posted @ 2014-03-14 14:18  hyx1  阅读(165)  评论(0编辑  收藏  举报