POJ - 3111 K Best （二分查找）
一、題目概述K Best
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
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 — 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 |
.[Submit] [Go Back] [Status] [Discuss]
二、題目釋義
有N顆珠寶，每顆珠寶的價值為vi，重量為wi。 女主不得已要賣掉部分珠寶，她想留下k顆珠寶，並要求(v1+v2+....vk) / (w1+w2+...wk)的值最大，輸出女主留下的珠寶的編號。（可不按輸入的順序輸出）。
三、思路分析
四、AC代碼
#include <iostream> #include <cstdio> #include <algorithm> #include <cmath> using namespace std; const int N = 1e5+5; const double esp = 1e-6; struct jewel { double v,w,t; int id; }S[N]; int n,k; bool cmp(jewel a, jewel b) { return a.t > b.t; } bool check(double x) { double sum = 0; for(int i=0; i<n; i++) { S[i].t = S[i].v - x*S[i].w; } sort(S,S+n,cmp); for(int i=0; i<k; i++) { sum += S[i].t; } return sum >= 0; } int main() { double l,r; double mid; while(scanf("%d%d",&n,&k) != EOF) { r = 0; for(int i=0; i<n; i++) { scanf("%lf%lf",&S[i].v,&S[i].w); S[i].id = i+1; r = max(r,S[i].v / S[i].w); } while(r-l>esp) { mid = (l+r) / 2; if(check(mid)) l = mid; else r = mid; } for(int i=0; i<k-1; i++) { cout << S[i].id << " "; } cout << S[k-1].id << endl; } return 0; }
POJ - 3111 K Best （二分查找）