uva 12124 - Assemble（二分定界，4级）

最新推荐文章于 2019-05-01 20:37:08 发布

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本文探讨了如何在预算内选择计算机组件以最大化最弱组件的质量，通过二分查找算法来确定最优价格区间，从而实现高效组装理想计算机。

Problem A - Assemble

Time limit: 2 seconds

Recently your team noticed that the computer you use to practice for programming contests is not good enough anymore. Therefore, you decide to buy a new computer.

To make the ideal computer for your needs, you decide to buy separate components and assemble the computer yourself. You need to buy exactly one of each type of component.

The problem is which components to buy. As you all know, the quality of a computer is equal to the quality of its weakest component. Therefore, you want to maximize the quality of the component with the lowest quality, while not exceeding your budget.

Input

On the first line one positive number: the number of testcases, at most 100. After that per testcase:

One line with two integers: 1 ≤ n ≤ 1000, the number of available components and 1 ≤ b ≤ 1000000000, your budget.
n lines in the following format: ``type name price quality'', where type is a string with the type of the component, name is a string with the unique name of the component, price is an integer (0 ≤ price < 1000000) which represents the price of the component and quality is an integer (0 ≤ quality ≤ 1000000000) which represents the quality of the component (higher is better). The strings contain only letters, digits and underscores and have a maximal length of 20 characters.

It will always possible to construct a computer with your budget.

Output

Per testcase:

One line with one integer: the maximal possible quality.

Sample Input

1
18 800
processor 3500_MHz 66 5
processor 4200_MHz 103 7
processor 5000_MHz 156 9
processor 6000_MHz 219 12
memory 1_GB 35 3
memory 2_GB 88 6
memory 4_GB 170 12
mainbord all_onboard 52 10
harddisk 250_GB 54 10
harddisk 500_FB 99 12
casing midi 36 10
monitor 17_inch 157 5
monitor 19_inch 175 7
monitor 20_inch 210 9
monitor 22_inch 293 12
mouse cordless_optical 18 12
mouse microsoft 30 9
keyboard office 4 10

Sample Output

The 2007 ACM Northwestern European Programming Contest

思路：二分定届判断，二分最小价值，看是否大于等于该价值的都选一个。满足条件里选最大

#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
#include<map>
#include<vector>
using namespace std;
const int mm=1e3+9;
const int oo=1e9;
class node
{
  public:int v,c;
};
vector<node>com[mm];
map<string,int>mp;
int n,budget,cnt;
int ID(string s)
{
  if(mp[s])return mp[s];
  else mp[s]=++cnt;
  return mp[s];
}
bool ok(int x)///判断
{
  int sum=0;
  for(int i=1;i<=cnt;++i)
  {
    int cheap=oo,z=com[i].size();
    for(int j=0;j<z;++j)
      if(com[i][j].v>=x)
      cheap=std::min(cheap,com[i][j].c);
    if(cheap==oo)return 0;
    sum+=cheap;
    if(sum>budget)return 0;
  }
  return 1;
}
int main()
{
  int cas,id;string s;node t;
  while(~scanf("%d",&cas))
  {
    while(cas--)
    { cnt=0;memset(com,0,sizeof(com));
      scanf("%d%d",&n,&budget);
      for(int i=0;i<n;++i)
      {cin>>s;id=ID(s);cin>>s;
      scanf("%d%d",&t.c,&t.v);
      com[id].push_back(t);
      }
      int l=0,r=1e9+3,mid,ans=0;
      while(l<=r)///二分最小价值
      {
        mid=(l+r)/2;
        if(ok(mid))l=mid+1,ans=std::max(ans,mid);
        else r=mid-1;
      }
      printf("%d\n",ans);
    }
  }
}