Inviting Friends

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 241 Accepted Submission(s): 97

Problem Description

You want to hold a birthday party, inviting as many friends as possible, but you have to prepare enough food for them. For each person, you need n kinds of ingredient to make good food. You can use the ingredients in your kitchen, or buy some new ingredient packages. There are exactly two kinds of packages for each kind of ingredient: small and large.

We use 6 integers to describe each ingredient: x, y, s1, p1, s2, p2, where x is the amount (of this ingredient) needed for one person, y is the amount currently available in the kitchen, s1 and p1 are the size (the amount of this ingredient in each package) and price of small packages, s2 and p2 are the size and price of large packages.

Given the amount of money you can spend, your task is to find the largest number of person who can serve. Note that you cannot buy only part of a package.

Input

There are at most 10 test cases. Each case begins with two integers n and m (1<=n<=100, 1<=m<=100000), the number of kinds of ingredient, and the amount of money you have. Each of the following n lines contains 6 positive integers x, y, s1, p1, s2, p2 to describe one kind of ingredient (10<=x<=100, 1<=y<=100, 1<=s1<=100, 10<=p1<=100, s1 s2<=100, p1p2<=100). The input ends with n = m = 0.

Output

For each test case, print the maximal number of people you can serve.

Sample Input

2 100

10 8 10 10 13 11

12 20 6 10 17 24

3 65

10 5 7 10 13 14

10 5 8 11 14 15

10 5 9 12 15 16

0 0

Sample Output

5

2

Source

2009 “NIT Cup” National Invitational Contest

二分+完全背包

由于不好计算具体的人的数量,可以提前估计好人的数量,采用二分的方式进行寻找答案

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <queue>

#include <map>

#include <algorithm>

#define INF 0x3f3f3f3f

using namespace std;

typedef long long LL;

const int MAX = 1e5+10;

int n,m;

int L,R;

int Dp[800000];

struct node

{

    int x;

    int y;

    int s1;

    int p1;

    int s2;

    int p2;

}Th[110];

int w[3],v[3];

int  Judge()//估计人的数量范围

{

    int tmp=INF;

    for(int i=1;i<=n;i++)

    {

        if(m*1.0/Th[i].p1*Th[i].s1>m*1.0/Th[i].p2*Th[i].s2)

        {

            tmp=min(tmp,(m/Th[i].p1*Th[i].s1+Th[i].y)/Th[i].x);

        }

        else

        {

            tmp=min(tmp,(m/Th[i].p2*Th[i].s2+Th[i].y)/Th[i].x);

        }

    }

    return tmp+10;

}

int Backpack(int s,int need)//完全背包

{

    for(int i=1;i<=need+Th[s].s2;i++)

    {

        Dp[i]=INF;

    }

    Dp[0]=0;

    int tmp=need+Th[s].s2;

    w[0]=Th[s].p1;

    w[1]=Th[s].p2;

    v[0]=Th[s].s1;

    v[1]=Th[s].s2;

    for(int i=0;i<2;i++)

    {

        for(int j=v[i];j<=tmp;j++)

        {

            Dp[j]=min(Dp[j-v[i]]+w[i],Dp[j]);

        }

    }

    int Max=INF;

    for(int i=need;i<=tmp;i++)

    {

        Max=min(Max,Dp[i]);

    }

    return Max;

}

bool BB(int num)

{

    int sum=0;

    for(int i=1;i<=n;i++)

    {

        int tmp=num*Th[i].x-Th[i].y;

        if(tmp<=0)

        {

            continue;

        }

        sum+=Backpack(i,tmp);

        if(sum>m)

        {

            return false;

        }

    }

    return true;

}

int main()

{

    while(scanf("%d %d",&n,&m)&&(n||m))

    {

        for(int i=1;i<=n;i++)

        {

            scanf("%d %d %d %d %d %d",&Th[i].x,&Th[i].y,&Th[i].s1,&Th[i].p1,&Th[i].s2,&Th[i].p2);

        }

        L=1;

        R=Judge();

        int ans=0;

        while(L<=R)

        {

            int mid=(L+R)>>1;

            if(BB(mid))

            {

                ans=max(ans,mid);

                L=mid+1;

            }

            else

            {

                R=mid-1;

            }

        }

        printf("%d\n",ans);

    }

    return 0;

}