POJ 2976 Dropping tests（普通01分数规划）

最新推荐文章于 2020-10-23 10:52:38 发布

glq007

最新推荐文章于 2020-10-23 10:52:38 发布

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分类专栏： 01分数规划文章标签： 01分数规划

本文链接：https://blog.youkuaiyun.com/glqac/article/details/41532911

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本文提供了一个使用二分法和Dinkelbach算法解决特定问题的示例代码。通过这两种方法求解最大性价比问题，展示了算法的具体实现过程。

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就是一道裸的题目，刚开始学试了2种写法，具体的方法学习可以看http://www.cnblogs.com/perseawe/archive/2012/05/03/01fsgh.html

二分代码：

//#pragma comment(linker, "/STACK:102400000,102400000")
#include<cstdio>
#include<ctype.h>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<cstdlib>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#include<ctime>
#include<string.h>
#include<string>
#include<sstream>
#include<bitset>
using namespace std;
#define ll long long
#define ull unsigned long long
#define eps 1e-7
#define NMAX 1000005
#define MOD 1000000007
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1)
template<class T>
inline void scan_d(T &ret)
{
    char c;
    int flag = 0;
    ret=0;
    while(((c=getchar())<'0'||c>'9')&&c!='-');
    if(c == '-')
    {
        flag = 1;
        c = getchar();
    }
    while(c>='0'&&c<='9') ret=ret*10+(c-'0'),c=getchar();
    if(flag) ret = -ret;
}
ll a[1005],b[1005];
double d[1005];
int main()
{
#ifdef GLQ
    freopen("input.txt","r",stdin);
//    freopen("o4.txt","w",stdout);
#endif // GLQ
    int n,k;
    while(~scanf("%d%d",&n,&k) && n+k)
    {
        for(int i = 1; i <= n; i++) scanf("%I64d",&a[i]);
        for(int i = 1; i <= n; i++) scanf("%I64d",&b[i]);
        double l = 0,r = 1,mid;
        while(abs(r-l) > eps)
        {
            mid = (l+r)/2;
            for(int i = 1; i <= n; i++) d[i] = a[i]-mid*b[i];
            sort(d+1,d+1+n);
            double tmp=0;
            for(int i = k+1; i <= n; i++) tmp += d[i];
//            cout<<tmp<<" "<<l<<" "<<r<<" "<<mid<<endl;
            if(tmp >= 0) l = mid;
            else r = mid;
        }
        int ans = l*100.0+0.5;
        printf("%d\n",ans);
    }
    return 0;
}

Dinkelbach写法：

//#pragma comment(linker, "/STACK:102400000,102400000")
#include<cstdio>
#include<ctype.h>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<cstdlib>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#include<ctime>
#include<string.h>
#include<string>
#include<sstream>
#include<bitset>
using namespace std;
#define ll long long
#define ull unsigned long long
#define eps 1e-7
#define NMAX 1000005
#define MOD 1000000007
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1)
template<class T>
inline void scan_d(T &ret)
{
    char c;
    int flag = 0;
    ret=0;
    while(((c=getchar())<'0'||c>'9')&&c!='-');
    if(c == '-')
    {
        flag = 1;
        c = getchar();
    }
    while(c>='0'&&c<='9') ret=ret*10+(c-'0'),c=getchar();
    if(flag) ret = -ret;
}
ll a[1005],b[1005];
struct node
{
    double ha;
    int pos;
    bool operator < (const node &t) const
    {
        return ha < t.ha;
    }
};
node d[1005];
int main()
{
#ifdef GLQ
    freopen("input.txt","r",stdin);
//    freopen("o4.txt","w",stdout);
#endif // GLQ
    int n,k;
    while(~scanf("%d%d",&n,&k) && n+k)
    {
        for(int i = 1; i <= n; i++) scanf("%I64d",&a[i]);
        for(int i = 1; i <= n; i++) scanf("%I64d",&b[i]);
        double ans=100,L = 0;
        ll p,q;
        while(abs(L-ans) > eps)
        {
            ans = L;
            for(int i = 1; i <= n; i++)
            {
                d[i].ha = a[i]-L*b[i]/100.0;
                d[i].pos = i;
            }
            sort(d+1,d+1+n);
            p = q = 0;
            for(int i = k+1; i <= n; i++)
            {
                p += a[d[i].pos];
                q += b[d[i].pos];
            }
            L = (double)p/q*100.0;
        }
        printf("%.0lf\n",ans);
    }
    return 0;
}