POJ 3111 K Best（01分数规划）

K Best

Time Limit: 8000MS		Memory Limit: 65536K
Total Submissions: 9876		Accepted: 2535
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

题目链接：POJ 3111

这题多了一个得输出方案序号，用结构体记录一下id就行，代码的话迭代或者二分都可以，但是迭代速度快很多很多

迭代代码：

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <sstream>
#include <numeric>
#include <cstring>
#include <bitset>
#include <string>
#include <deque>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 100010;
const double eps = 1e-9;
struct info
{
    double w, v, d;
    int id;
    bool operator<(const info &rhs)const
    {
        return d > rhs.d;
    }
};
info A[N];

double getnewk(int n, int k, double r)
{
    double V = 0.0, W = 0.0;
    for (int i = 0; i < n; ++i)
        A[i].d = A[i].v - r * A[i].w;
    sort(A, A + n);
    for (int i = 0; i < k; ++i)
    {
        V += A[i].v;
        W += A[i].w;
    }
    return V / W;
}
int main(void)
{
    int n, k, i;
    while (~scanf("%d%d", &n, &k))
    {
        for (i = 0; i < n; ++i)
        {
            scanf("%lf%lf", &A[i].v, &A[i].w);
            A[i].id = i + 1;
        }
        double ans = 1, temp = 1;
        while (1)
        {
            temp = getnewk(n, k, ans);
            if (fabs(temp - ans) < eps)
                break;
            ans = temp;
        }
        for (i = 0; i < k; ++i)
            printf("%d%s", A[i].id, i == k - 1 ? "
" : " ");
    }
    return 0;
}