POJ3111-K Best

K Best

Time Limit: 8000MS Memory Limit: 65536KTotal Submissions: 10263 Accepted: 2644Case 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 21 11 21 3

Sample Output

1 2

Source

题意：有n件珠宝，每样珠宝有两个属性，价值v和重量w。选择k件，使得k件的价值和除以k件的重量和最大，也就是单位重量的价值最大

解题思路：二分比值，和POJ2976-Dropping tests一样

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <cmath>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int n,k;struct node{int a, b,id;double p;}x[100009];bool cmp(node a, node b){return a.p > b.p;}bool check(double xx){for (int i = 1; i <= n; i++) x[i].p = 1.0*x[i].a -1.0* x[i].b*xx;sort(x + 1, x + 1 + n, cmp);double sum1 = 0, sum2 = 0;for (int i = 1; i <= k; i++) sum1 += x[i].a, sum2 += x[i].b;return sum1 / sum2 > xx;}int main(){while (~scanf("%d%d", &n, &k)){for (int i = 1; i <= n; i++) scanf("%d%d", &x[i].a, &x[i].b), x[i].id = i;double l = 0, r = 1.0*INF;while (fabs(r - l) > 1e-8){double mid = (l + r) / 2;if (check(mid)) l = mid;else r = mid;}printf("%d", x[1].id);for (int i = 2; i <= k; i++) printf(" %d", x[i].id);printf("\n");}return 0;}