POJ3111 K Best —— 01分数规划 二分法

题目链接：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

 

 

 

 

题解：

 http://www.cnblogs.com/DOLFAMINGO/p/7563213.html

 

 

代码一：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
#define ms(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long LL;
const double EPS = 1e-8;
const int INF = 2e9;
const LL LNF = 2e18;
const int MAXN = 1e5+10;

struct node
{
    double d;
    int a, b, id;
    bool operator<(const node x)const{ 
        return d>x.d;
    }
}q[MAXN];
int n, k;

bool test(double L)
{
    for(int i = 1; i<=n; i++)
        q[i].d = 1.0*q[i].a - L*q[i].b;

    sort(q+1, q+1+n);
    double sum = 0;
    for(int i = 1; i<=k; i++)   //取前k大的数
        sum += q[i].d;
    return sum>=0;
}

int main()
{
    while(scanf("%d%d", &n, &k)!=EOF)
    {
         //一次性把所有信息都录入结构体中，当排序时，即使打乱了顺序，仍然还记得初始下标。
        for(int i = 1; i<=n; i++)  
        {
            scanf("%d%d", &q[i].a, &q[i].b);
            q[i].id = i;    
        }

        double l = 0, r = 1e7;
        while(l+EPS<=r)
        {
            double mid = (l+r)/2;
            if(test(mid))
                l = mid + EPS;
            else
                r = mid - EPS;
        }

        for(int i = 1; i<=k; i++)
            printf("%d ", q[i].id);
        printf("
");
    }
}

 

代码二：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
#define ms(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long LL;
const double EPS = 1e-8;
const int INF = 2e9;
const LL LNF = 2e18;
const int MAXN = 1e5+10;

struct node
{
    double d;
    int id;
    bool operator<(const node x)const{
        return d>x.d;
    }
}q[MAXN];

int n, k;
int a[MAXN], b[MAXN];

bool test(double L)
{
    for(int i = 1; i<=n; i++)   //每一次q[i]都重新更新，与a[i],b[i]独立开来
    {
        q[i].id = i;
        q[i].d = 1.0*a[i] - L*b[i];
    }
    sort(q+1, q+1+n);
    double sum = 0;
    for(int i = 1; i<=k; i++)
        sum += q[i].d;
    return sum>=0;
}

int main()
{
    while(scanf("%d%d", &n, &k)!=EOF)
    {
        for(int i = 1; i<=n; i++)
            scanf("%d%d", &a[i], &b[i]);

        double l = 0, r = 1e7;
        while(l+EPS<=r)
        {
            double mid = (l+r)/2;
            if(test(mid))
                l = mid + EPS;
            else
                r = mid - EPS;
        }

        for(int i = 1; i<=k; i++)
            printf("%d ", q[i].id);
        printf("
");
    }
}

错误代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
#define ms(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long LL;
const double EPS = 1e-8;
const int INF = 2e9;
const LL LNF = 2e18;
const int MAXN = 1e5+10;

struct node
{
    double d;
    int id;
    bool operator<(const node x)const{
        return d>x.d;
    }
}q[MAXN];

int n, k;
int a[MAXN], b[MAXN], ans[MAXN];

bool test(double L)
{
    //经过一次排序后，q[i].id不再等于i，所以出错。应该同时更新q[i].id
    for(int i = 1; i<=n; i++)
        q[i].d = 1.0*a[i] - L*b[i];

    sort(q+1, q+1+n);
    double sum = 0;
    for(int i = 1; i<=k; i++)
        sum += q[i].d;
    return sum>0;
}

int main()
{
    while(scanf("%d%d", &n, &k)!=EOF)
    {
        for(int i = 1; i<=n; i++)
        {
            scanf("%d%d", &a[i], &b[i]);
            q[i].id = i;       
        }

        double l = 0, r = 1e7;
        while(l+EPS<=r)
        {
            double mid = (l+r)/2;
            if(test(mid))
                l = mid + EPS;
            else
                r = mid - EPS;
        }

        for(int i = 1; i<=k; i++)
            printf("%d ", q[i].id);
        printf("
");
    }
}