題目鏈接：http://poj.org/problem?id=3111

?

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

3 2 1 1 1 2 1 3

Sample Output

1 2

Source

Northeastern Europe 2005, Northern Subregion 題解： http://www.cnblogs.com/DOLFAMINGO/p/7563213.html 代碼一： 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大的數(shù) 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錄入結(jié)構(gòu)體中，當(dāng)排序時(shí)，即使打亂了順序，仍然還記得初始下標(biāo)。 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("\n"); 66 } 67 } View Code 代碼二： 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，與a[i],b[i]獨(dú)立開來(lái) 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("\n"); 66 } 67 } View Code

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錯(cuò)誤代碼：

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 //經(jīng)過(guò)一次排序后，q[i].id不再等于i，所以出錯(cuò)。應(yīng)該同時(shí)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("\n"); 68 } 69 } View Code

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轉(zhuǎn)載于:https://www.cnblogs.com/DOLFAMINGO/p/7571434.html

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