UVALive 3971

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;
}