poj 3111 K Best 二分

K Best

Time Limit: 8000MS Memory Limit: 65536KTotal Submissions: 11677 Accepted: 3016Case 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

题意：从n个宝石里选k个，求的最大值。

分析：经典题啊，公式少许变化，然后排序、二分。

#include <cstdio>#include <cstdlib>#include <iostream>#include <stack>#include <queue>#include <algorithm>#include <cstring>#include <string>#include <cmath>#include <vector>#include <bitset>#include <list>#include <sstream>#include <set>#include <functional>using namespace std;#define INF 0x3f3f3f3f#define MAX 100#define ESP 1e-6typedef long long ll;int n,k;struct node{int id,v,w;double ans;bool operator < (const node &a) const{return ans > a.ans;}}y[1000001];bool C(double x){for (int i = 0; i < n; i += 1) y[i].ans = y[i].v-x*y[i].w;sort(y,y+n);double sum = 0;for (int i = 0; i < k; i += 1) sum += y[i].ans;return sum >= 0;}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);    freopen("out.txt", "w", stdout);#endifcin >> n >> k;for (int i = 0; i < n; i += 1) {cin >> y[i].v >> y[i].w;y[i].id = i+1;}double lb = 0,ub = INF;while (ub-lb>=ESP){double mid = (ub+lb)/2;if(C(mid)) lb = mid;else ub = mid;}for (int i = 0; i < k-1; i += 1) printf("%d ",y[i].id);printf("%d\n",y[k-1].id);return 0;}