题意：有b块钱。想要组装一台电脑，给出n个配件的种类，名字，价格，品质因子。若各种类配件各买一个，总价格<=b，求最差品质配件的最大品质因子。
思路：
求最大的最小值一般用二分法。
在（0。maxq）内进行二分，判定q作为最差品质因子是否可行。
大白书原题。比較考验代码功底。
code：
/*
* @author Novicer
* language : C++/C
*/
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
using namespace std;
const double eps(1e-8);
typedef long long lint;
const int maxn = 1000 + 5;
map<string,int> id;
int n,b;
int cnt = 0;
struct Comp{
int price;
int quality;
};
vector<Comp> comp[maxn];
int ID(string s){
if(!id.count(s)){
id[s] = cnt++;
}
return id[s];
}
bool ok(int q){
int sum = 0;
// cout << q << endl;
for(int i = 0 ; i < cnt ; i++){
int cheapest = b + 1;
for(int j = 0 ; j < comp[i].size() ; j++){
// cout << comp[i][j].price << endl;
if(q <= comp[i][j].quality) cheapest = min(cheapest , comp[i][j].price);
}
if(cheapest == b+1) return false;
sum += cheapest;
// cout << "q : " << q << " sum : " << sum << endl;
if(sum > b) return false;
}
return true;
}
int solve(int l , int r){
while(l < r){
int m = l + (r - l + 1)/2;
if(ok(m)) l = m;
else r = m - 1;
// cout << m << endl;
}
return l;
}
int main(){
// freopen("input.txt","r",stdin);
int T;
cin >> T;
while(T--){
cnt = 0;
cin >> n >> b;
for(int i = 0 ; i < n ; i++) comp[i].clear();
int maxq = 0;
for(int i = 1 ; i <= n ; i++){
string type,name;
int p,q;
cin >> type >> name >> p >> q;
// cout << type << name << p << q << endl;
maxq = max(maxq , q);
Comp tmp;
tmp.price = p; tmp.quality = q;
comp[ID(type)].push_back(tmp);
}
// cout << maxq << endl;
int L = 0 , R = maxq;
int ans = solve(L,R);
cout << ans << endl;
}
return 0;
}
