LA3971 - Assemble

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 rst 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
9

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <map>
using namespace std;
const int MAX = 1000000010;
const int MAXN = 1100;
struct node{
    int id;
    int price;
    int val;
}WuPing[MAXN];
bool cmp(node a, node b)
{
    return a.val>b.val;
}
int n, money;
map<string,int>mID;
int id;
int sc[MAXN];
bool check(int mid)
{
    int i;
    for(i=1; i<id; i++) sc[i] = MAX;
    for(i=0; i<n; i++)
    {
        if(WuPing[i].val < mid)break;
        int x = WuPing[i].id;
        sc[x] = min(sc[x],WuPing[i].price);
    }
    long long s = money;
    for(i=1; i<id; i++)
    {
        if(sc[i]==MAX)return false;
        s-=sc[i];
    }
    return s>=0;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d",&n,&money);
        mID.clear();
        id = 1; 
        char Lei[50], Name[50];
        int p, v, i;
        for(i=0; i<n; i++)
        {
            scanf("%s%s%d%d",Lei,Name,&p,&v);
            if(mID[Lei])
            {
                WuPing[i].id = mID[Lei];
            }else{
                WuPing[i].id = id;
                mID[Lei] = id++;
            }
            WuPing[i].price = p;
            WuPing[i].val = v;
        }
        sort(WuPing,WuPing+n,cmp);
        int max1 = MAX, min1 = -1, mid;
        while(max1>min1+1)
        {
            mid = (max1+min1)/2;
            
            if(check(mid))
            {
                min1 = mid;
            }else{
                max1 = mid;
            }
        } 
        printf("%d\n",min1);
    }
    return 0;
}