POJ3111 K Best —— 01分数规划二分法

K Best

Time Limit: 8000MS		Memory Limit: 65536K
Total Submissions: 11380		Accepted: 2935
Case Time Limit: 2000MS		Special Judge

Description

Demy has n jewels. Each of her jewels has some value v_i and weight w_i.

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, …, i_k} 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 ≤ k ≤ n ≤ 100 000).

The following n lines contain two integer numbers each — v_i and w_i (0 ≤ v_i ≤ 10⁶, 1 ≤ w_i ≤ 10⁶, both the sum of all v_i and the sum of all w_i 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

Sample Output

1 2

Source

Northeastern Europe 2005, Northern Subregion

题解：

http://www.cnblogs.com/DOLFAMINGO/p/7563213.html

代码一：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <vector>
 7 #include <queue>
 8 #include <stack>
 9 #include <map>
10 #include <string>
11 #include <set>
12 #define ms(a,b) memset((a),(b),sizeof((a)))
13 using namespace std;
14 typedef long long LL;
15 const double EPS = 1e-8;
16 const int INF = 2e9;
17 const LL LNF = 2e18;
18 const int MAXN = 1e5+10;
19 
20 struct node
21 {
22     double d;
23     int a, b, id;
24     bool operator<(const node x)const{ 
25         return d>x.d;
26     }
27 }q[MAXN];
28 int n, k;
29 
30 bool test(double L)
31 {
32     for(int i = 1; i<=n; i++)
33         q[i].d = 1.0*q[i].a - L*q[i].b;
34 
35     sort(q+1, q+1+n);
36     double sum = 0;
37     for(int i = 1; i<=k; i++)   //取前k大的数
38         sum += q[i].d;
39     return sum>=0;
40 }
41 
42 int main()
43 {
44     while(scanf("%d%d", &n, &k)!=EOF)
45     {
46          //一次性把所有信息都录入结构体中，当排序时，即使打乱了顺序，仍然还记得初始下标。
47         for(int i = 1; i<=n; i++)  
48         {
49             scanf("%d%d", &q[i].a, &q[i].b);
50             q[i].id = i;    
51         }
52 
53         double l = 0, r = 1e7;
54         while(l+EPS<=r)
55         {
56             double mid = (l+r)/2;
57             if(test(mid))
58                 l = mid + EPS;
59             else
60                 r = mid - EPS;
61         }
62 
63         for(int i = 1; i<=k; i++)
64             printf("%d ", q[i].id);
65         printf("
");
66     }
67 }

View Code

代码二：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <vector>
 7 #include <queue>
 8 #include <stack>
 9 #include <map>
10 #include <string>
11 #include <set>
12 #define ms(a,b) memset((a),(b),sizeof((a)))
13 using namespace std;
14 typedef long long LL;
15 const double EPS = 1e-8;
16 const int INF = 2e9;
17 const LL LNF = 2e18;
18 const int MAXN = 1e5+10;
19 
20 struct node
21 {
22     double d;
23     int id;
24     bool operator<(const node x)const{
25         return d>x.d;
26     }
27 }q[MAXN];
28 
29 int n, k;
30 int a[MAXN], b[MAXN];
31 
32 bool test(double L)
33 {
34     for(int i = 1; i<=n; i++)   //每一次q[i]都重新更新，与a[i],b[i]独立开来
35     {
36         q[i].id = i;
37         q[i].d = 1.0*a[i] - L*b[i];
38     }
39     sort(q+1, q+1+n);
40     double sum = 0;
41     for(int i = 1; i<=k; i++)
42         sum += q[i].d;
43     return sum>=0;
44 }
45 
46 int main()
47 {
48     while(scanf("%d%d", &n, &k)!=EOF)
49     {
50         for(int i = 1; i<=n; i++)
51             scanf("%d%d", &a[i], &b[i]);
52 
53         double l = 0, r = 1e7;
54         while(l+EPS<=r)
55         {
56             double mid = (l+r)/2;
57             if(test(mid))
58                 l = mid + EPS;
59             else
60                 r = mid - EPS;
61         }
62 
63         for(int i = 1; i<=k; i++)
64             printf("%d ", q[i].id);
65         printf("
");
66     }
67 }

View Code

错误代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <vector>
 7 #include <queue>
 8 #include <stack>
 9 #include <map>
10 #include <string>
11 #include <set>
12 #define ms(a,b) memset((a),(b),sizeof((a)))
13 using namespace std;
14 typedef long long LL;
15 const double EPS = 1e-8;
16 const int INF = 2e9;
17 const LL LNF = 2e18;
18 const int MAXN = 1e5+10;
19 
20 struct node
21 {
22     double d;
23     int id;
24     bool operator<(const node x)const{
25         return d>x.d;
26     }
27 }q[MAXN];
28 
29 int n, k;
30 int a[MAXN], b[MAXN], ans[MAXN];
31 
32 bool test(double L)
33 {
34     //经过一次排序后，q[i].id不再等于i，所以出错。应该同时更新q[i].id
35     for(int i = 1; i<=n; i++)
36         q[i].d = 1.0*a[i] - L*b[i];
37 
38     sort(q+1, q+1+n);
39     double sum = 0;
40     for(int i = 1; i<=k; i++)
41         sum += q[i].d;
42     return sum>0;
43 }
44 
45 int main()
46 {
47     while(scanf("%d%d", &n, &k)!=EOF)
48     {
49         for(int i = 1; i<=n; i++)
50         {
51             scanf("%d%d", &a[i], &b[i]);
52             q[i].id = i;       
53         }
54 
55         double l = 0, r = 1e7;
56         while(l+EPS<=r)
57         {
58             double mid = (l+r)/2;
59             if(test(mid))
60                 l = mid + EPS;
61             else
62                 r = mid - EPS;
63         }
64 
65         for(int i = 1; i<=k; i++)
66             printf("%d ", q[i].id);
67         printf("
");
68     }
69 }

View Code

POJ3111 K Best —— 01分数规划 二分法

POJ3111 K Best —— 01分数规划二分法